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| hi everyone welcome back so today I'd |
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| align:start position:0% |
| hi everyone welcome back so today I'd |
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| align:start position:0% |
| hi everyone welcome back so today I'd |
| like to take a look at solving a linear |
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| align:start position:0% |
| like to take a look at solving a linear |
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| align:start position:0% |
| like to take a look at solving a linear |
| o de but with a periodic input and |
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| o de but with a periodic input and |
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| align:start position:0% |
| o de but with a periodic input and |
| specifically we're asked to find one |
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| specifically we're asked to find one |
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| align:start position:0% |
| specifically we're asked to find one |
| solution that a particular solution |
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| solution that a particular solution |
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| solution that a particular solution |
| which is also the periodic solution of |
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| which is also the periodic solution of |
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| align:start position:0% |
| which is also the periodic solution of |
| the differential equation X dot plus two |
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| the differential equation X dot plus two |
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| align:start position:0% |
| the differential equation X dot plus two |
| x dot plus 4x equals |
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| align:start position:0% |
| x dot plus 4x equals |
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| align:start position:0% |
| x dot plus 4x equals |
| the square wave function so the square |
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| the square wave function so the square |
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| the square wave function so the square |
| wave function is periodic function with |
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| wave function is periodic function with |
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| align:start position:0% |
| wave function is periodic function with |
| period 2pi it's defined as minus 1 and 1 |
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| period 2pi it's defined as minus 1 and 1 |
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| align:start position:0% |
| period 2pi it's defined as minus 1 and 1 |
| on the intervals minus prior to 0 and 0 |
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| on the intervals minus prior to 0 and 0 |
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| align:start position:0% |
| on the intervals minus prior to 0 and 0 |
| to PI and we know that the square wave |
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| align:start position:0% |
| to PI and we know that the square wave |
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| align:start position:0% |
| to PI and we know that the square wave |
| function has the following Fourier |
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| function has the following Fourier |
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| function has the following Fourier |
| series |
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| series |
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| series |
| so let you think about this problem for |
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| align:start position:0% |
| so let you think about this problem for |
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| align:start position:0% |
| so let you think about this problem for |
| a moment and I'll come back in a second |
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| align:start position:0% |
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| hi everyone welcome back |
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| align:start position:0% |
| hi everyone welcome back |
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| align:start position:0% |
| hi everyone welcome back |
| okay so the reason we've been studying |
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| align:start position:0% |
| okay so the reason we've been studying |
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| align:start position:0% |
| okay so the reason we've been studying |
| Fourier series is to essentially solve |
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| align:start position:0% |
| Fourier series is to essentially solve |
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| Fourier series is to essentially solve |
| differential equations with complicated |
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| differential equations with complicated |
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| differential equations with complicated |
| forcing inputs on the right hand side |
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| forcing inputs on the right hand side |
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| forcing inputs on the right hand side |
| which are periodic and the reason we've |
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| align:start position:0% |
| which are periodic and the reason we've |
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| align:start position:0% |
| which are periodic and the reason we've |
| been studying Fourier series is because |
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| been studying Fourier series is because |
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| been studying Fourier series is because |
| we know that differential equations with |
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| we know that differential equations with |
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| we know that differential equations with |
| sines and cosines as forcing terms on |
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| sines and cosines as forcing terms on |
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| sines and cosines as forcing terms on |
| the right-hand side are relatively easy |
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| the right-hand side are relatively easy |
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| the right-hand side are relatively easy |
| to solve and |
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| to solve and |
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| to solve and |
| we want to be able to solve the same |
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| we want to be able to solve the same |
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| align:start position:0% |
| we want to be able to solve the same |
| differential equation with a more |
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| differential equation with a more |
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| align:start position:0% |
| differential equation with a more |
| complicated periodic function on the |
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| complicated periodic function on the |
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| complicated periodic function on the |
| right hand side so the general approach |
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| right hand side so the general approach |
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| right hand side so the general approach |
| is to first |
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| do you compose |
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| the right-hand side |
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| into a Fourier series |
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| okay and this step is essentially |
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| align:start position:0% |
| okay and this step is essentially |
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| align:start position:0% |
| okay and this step is essentially |
| already done for us we're told what the |
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| align:start position:0% |
| already done for us we're told what the |
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| align:start position:0% |
| already done for us we're told what the |
| Fourier series is |
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| Fourier series is |
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| Fourier series is |
| and what we do is we solve |
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| align:start position:0% |
| and what we do is we solve |
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| align:start position:0% |
| and what we do is we solve |
| the Eau de |
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| the Eau de |
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| the Eau de |
| X dot dot plus 2 X dot plus 4x and |
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| X dot dot plus 2 X dot plus 4x and |
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| align:start position:0% |
| X dot dot plus 2 X dot plus 4x and |
| I'm just going to take one |
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| align:start position:0% |
| I'm just going to take one |
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| align:start position:0% |
| I'm just going to take one |
| term of the Fourier series sine NT |
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| align:start position:0% |
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| so if we know the right hand side is a |
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| align:start position:0% |
| so if we know the right hand side is a |
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| align:start position:0% |
| so if we know the right hand side is a |
| sum of a whole bunch of signs what we're |
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| align:start position:0% |
| sum of a whole bunch of signs what we're |
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| align:start position:0% |
| sum of a whole bunch of signs what we're |
| going to do is we're going to solve for |
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| align:start position:0% |
| going to do is we're going to solve for |
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| align:start position:0% |
| going to do is we're going to solve for |
| any specific one of those signs so we |
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| align:start position:0% |
| any specific one of those signs so we |
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| align:start position:0% |
| any specific one of those signs so we |
| want to solve this differential equation |
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| align:start position:0% |
| want to solve this differential equation |
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| align:start position:0% |
| want to solve this differential equation |
| and |
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| align:start position:0% |
| and |
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| align:start position:0% |
| and |
| then we use |
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| superposition |
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| align:start position:0% |
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| so if we know what the solution is to 1 |
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| align:start position:0% |
| so if we know what the solution is to 1 |
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| align:start position:0% |
| so if we know what the solution is to 1 |
| sine NT and we know that the right hand |
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| align:start position:0% |
| sine NT and we know that the right hand |
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| align:start position:0% |
| sine NT and we know that the right hand |
| side forcing is a sum of many sign |
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| align:start position:0% |
| side forcing is a sum of many sign |
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| align:start position:0% |
| side forcing is a sum of many sign |
| entities with appropriate weight factors |
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| entities with appropriate weight factors |
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| entities with appropriate weight factors |
| then we can use superposition to |
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| then we can use superposition to |
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| align:start position:0% |
| then we can use superposition to |
| construct |
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| a |
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| final solution |
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| align:start position:0% |
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| okay so this is the method to attack |
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| align:start position:0% |
| okay so this is the method to attack |
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| align:start position:0% |
| okay so this is the method to attack |
| this problem |
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| this problem |
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| align:start position:0% |
| this problem |
| alright so we've already done step 1 or |
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| align:start position:0% |
| alright so we've already done step 1 or |
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| align:start position:0% |
| alright so we've already done step 1 or |
| we were given step 1 and we wanted to |
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| align:start position:0% |
| we were given step 1 and we wanted to |
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| align:start position:0% |
| we were given step 1 and we wanted to |
| solve step 2 and we can solve this just |
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| align:start position:0% |
| solve step 2 and we can solve this just |
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| align:start position:0% |
| solve step 2 and we can solve this just |
| using the exponential response formula |
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| align:start position:0% |
| using the exponential response formula |
| |
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| align:start position:0% |
| using the exponential response formula |
| so what we're going to do is we're just |
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| align:start position:0% |
| so what we're going to do is we're just |
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| align:start position:0% |
| so what we're going to do is we're just |
| going to |
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| going to |
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| going to |
| complexity the right hand side |
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| align:start position:0% |
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| align:start position:0% |
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| so if I want to solve the differential |
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| align:start position:0% |
| so if I want to solve the differential |
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| align:start position:0% |
| so if I want to solve the differential |
| equation x + 2 X dot plus 4x equals sign |
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| align:start position:0% |
| equation x + 2 X dot plus 4x equals sign |
| |
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| align:start position:0% |
| equation x + 2 X dot plus 4x equals sign |
| in T I'm going to denote the solution |
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| align:start position:0% |
| in T I'm going to denote the solution |
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| align:start position:0% |
| in T I'm going to denote the solution |
| with a subscript n and it's going to be |
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| align:start position:0% |
| with a subscript n and it's going to be |
| |
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| align:start position:0% |
| with a subscript n and it's going to be |
| the imaginary part of |
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| align:start position:0% |
| the imaginary part of |
| |
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| align:start position:0% |
| the imaginary part of |
| and I'm just using the exponential |
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| align:start position:0% |
| and I'm just using the exponential |
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| align:start position:0% |
| and I'm just using the exponential |
| response formula 1 over the |
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| align:start position:0% |
| response formula 1 over the |
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| align:start position:0% |
| response formula 1 over the |
| characteristic polynomial evaluated at I |
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| align:start position:0% |
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| ne2 the int |
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| align:start position:0% |
| ne2 the int |
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| align:start position:0% |
| ne2 the int |
| you might ask how did I get that |
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| align:start position:0% |
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| well I basically just took the |
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| align:start position:0% |
| well I basically just took the |
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| align:start position:0% |
| well I basically just took the |
| complexity aight formula or sorry the |
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| align:start position:0% |
| complexity aight formula or sorry the |
| |
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| align:start position:0% |
| complexity aight formula or sorry the |
| complex if I'd equation |
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| align:start position:0% |
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| align:start position:0% |
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| and noted that sine NT was the imaginary |
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| align:start position:0% |
| and noted that sine NT was the imaginary |
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| align:start position:0% |
| and noted that sine NT was the imaginary |
| part of e to the int and then use the |
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| align:start position:0% |
| part of e to the int and then use the |
| |
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| align:start position:0% |
| part of e to the int and then use the |
| exponential response formula and I want |
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| align:start position:0% |
| exponential response formula and I want |
| |
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| align:start position:0% |
| exponential response formula and I want |
| to take the imaginary part at the end of |
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| align:start position:0% |
| to take the imaginary part at the end of |
| |
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| align:start position:0% |
| to take the imaginary part at the end of |
| the day |
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| align:start position:0% |
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| align:start position:0% |
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| so what is a the characteristic |
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| align:start position:0% |
| so what is a the characteristic |
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| align:start position:0% |
| so what is a the characteristic |
| polynomial in this case it's s squared |
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| align:start position:0% |
| polynomial in this case it's s squared |
| |
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| align:start position:0% |
| polynomial in this case it's s squared |
| plus 2 s plus 4 |
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| plus 2 s plus 4 |
| |
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| align:start position:0% |
| plus 2 s plus 4 |
| which means that P of I n |
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| align:start position:0% |
| which means that P of I n |
| |
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| align:start position:0% |
| which means that P of I n |
| is going to be negative n squared plus 4 |
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| align:start position:0% |
| is going to be negative n squared plus 4 |
| |
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| align:start position:0% |
| is going to be negative n squared plus 4 |
| so I'll just group the real terms |
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| align:start position:0% |
| so I'll just group the real terms |
| |
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| align:start position:0% |
| so I'll just group the real terms |
| together plus 2i n |
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| align:start position:0% |
| together plus 2i n |
| |
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| align:start position:0% |
| together plus 2i n |
| so the 2i n comes from the 2's term |
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| align:start position:0% |
| so the 2i n comes from the 2's term |
| |
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| align:start position:0% |
| so the 2i n comes from the 2's term |
| and |
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| align:start position:0% |
| and |
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| align:start position:0% |
| and |
| then X of n is going to be the imaginary |
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| align:start position:0% |
| then X of n is going to be the imaginary |
| |
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| align:start position:0% |
| then X of n is going to be the imaginary |
| part of |
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| align:start position:0% |
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| align:start position:0% |
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| 1 over 4 minus n squared |
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| align:start position:0% |
| 1 over 4 minus n squared |
| |
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| align:start position:0% |
| 1 over 4 minus n squared |
| plus 2i |
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| align:start position:0% |
| plus 2i |
| |
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| align:start position:0% |
| plus 2i |
| ne2 the int |
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| align:start position:0% |
| ne2 the int |
| |
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| align:start position:0% |
| ne2 the int |
| and |
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| align:start position:0% |
| and |
| |
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| align:start position:0% |
| and |
| I'm going to use the amplify |
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| align:start position:0% |
| I'm going to use the amplify |
| |
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| align:start position:0% |
| I'm going to use the amplify |
| the amplitude phase form to convert this |
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| align:start position:0% |
| the amplitude phase form to convert this |
| |
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| align:start position:0% |
| the amplitude phase form to convert this |
| or to convert this Cartesian complex |
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| align:start position:0% |
| or to convert this Cartesian complex |
| |
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| align:start position:0% |
| or to convert this Cartesian complex |
| number |
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| align:start position:0% |
| number |
| |
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| align:start position:0% |
| number |
| sorry I'm going to convert this |
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| align:start position:0% |
| sorry I'm going to convert this |
| |
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| align:start position:0% |
| sorry I'm going to convert this |
| Cartesian complex number into an |
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| align:start position:0% |
| Cartesian complex number into an |
| |
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| align:start position:0% |
| Cartesian complex number into an |
| amplitude phase form just because it's |
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| align:start position:0% |
| amplitude phase form just because it's |
| |
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| align:start position:0% |
| amplitude phase form just because it's |
| going to make taking the imaginary part |
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| align:start position:0% |
| going to make taking the imaginary part |
| |
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| align:start position:0% |
| going to make taking the imaginary part |
| of the solution very easy at the end of |
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| align:start position:0% |
| of the solution very easy at the end of |
| |
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| align:start position:0% |
| of the solution very easy at the end of |
| the day |
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| align:start position:0% |
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| align:start position:0% |
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| so the amplitude of this complex number |
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| align:start position:0% |
| so the amplitude of this complex number |
| |
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| align:start position:0% |
| so the amplitude of this complex number |
| is for n sorry 4 minus N squared |
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| align:start position:0% |
| is for n sorry 4 minus N squared |
| |
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| align:start position:0% |
| is for n sorry 4 minus N squared |
| quantity squared |
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| align:start position:0% |
| quantity squared |
| |
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| align:start position:0% |
| quantity squared |
| plus the imaginary part squared so |
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| align:start position:0% |
| plus the imaginary part squared so |
| |
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| align:start position:0% |
| plus the imaginary part squared so |
| that's 4n squared |
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| align:start position:0% |
| that's 4n squared |
| |
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| align:start position:0% |
| that's 4n squared |
| square rooted and |
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| align:start position:0% |
| square rooted and |
| |
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| align:start position:0% |
| square rooted and |
| then we have e |
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| align:start position:0% |
| then we have e |
| |
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| align:start position:0% |
| then we have e |
| upstairs e |
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| align:start position:0% |
| upstairs e |
| |
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| align:start position:0% |
| upstairs e |
| int and downstairs is going to be e I n |
|
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| align:start position:0% |
| int and downstairs is going to be e I n |
| |
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|
| align:start position:0% |
| int and downstairs is going to be e I n |
| Phi sorry i phi and |
|
|
| align:start position:0% |
| Phi sorry i phi and |
| |
|
|
| align:start position:0% |
| Phi sorry i phi and |
| i'm going to |
|
|
| align:start position:0% |
| i'm going to |
| |
|
|
| align:start position:0% |
| i'm going to |
| put a subscript n on the phi just |
|
|
| align:start position:0% |
| put a subscript n on the phi just |
| |
|
|
| align:start position:0% |
| put a subscript n on the phi just |
| because for each |
|
|
| align:start position:0% |
| because for each |
| |
|
|
| align:start position:0% |
| because for each |
| for each complex number we're going to |
|
|
| align:start position:0% |
| for each complex number we're going to |
| |
|
|
| align:start position:0% |
| for each complex number we're going to |
| have a different phase Phi and that |
|
|
| align:start position:0% |
| have a different phase Phi and that |
| |
|
|
| align:start position:0% |
| have a different phase Phi and that |
| phase Phi n is |
|
|
| align:start position:0% |
| phase Phi n is |
| |
|
|
| align:start position:0% |
| phase Phi n is |
| going to be the arctangent of |
|
|
| align:start position:0% |
| going to be the arctangent of |
| |
|
|
| align:start position:0% |
| going to be the arctangent of |
| 2 n |
|
|
| align:start position:0% |
| 2 n |
| |
|
|
| align:start position:0% |
| 2 n |
| divided divided by 4 minus n squared |
|
|
| align:start position:0% |
| |
| |
|
|
| align:start position:0% |
| |
| like that |
|
|
| align:start position:0% |
| |
| |
|
|
| align:start position:0% |
| |
| okay so now i can combine phi then with |
|
|
| align:start position:0% |
| okay so now i can combine phi then with |
| |
|
|
| align:start position:0% |
| okay so now i can combine phi then with |
| the upstairs term int in the exponent |
|
|
| align:start position:0% |
| the upstairs term int in the exponent |
| |
|
|
| align:start position:0% |
| the upstairs term int in the exponent |
| and when I take the imaginary part I'm |
|
|
| align:start position:0% |
| and when I take the imaginary part I'm |
| |
|
|
| align:start position:0% |
| and when I take the imaginary part I'm |
| only going to be left with sine of n t |
|
|
| align:start position:0% |
| only going to be left with sine of n t |
| |
|
|
| align:start position:0% |
| only going to be left with sine of n t |
| minus Phi n |
|
|
| align:start position:0% |
| minus Phi n |
| |
|
|
| align:start position:0% |
| minus Phi n |
| so X event |
|
|
| align:start position:0% |
| so X event |
| |
|
|
| align:start position:0% |
| so X event |
| yep going is going to be 1 over |
|
|
| align:start position:0% |
| yep going is going to be 1 over |
| |
|
|
| align:start position:0% |
| yep going is going to be 1 over |
| 4 minus N squared |
|
|
| align:start position:0% |
| 4 minus N squared |
| |
|
|
| align:start position:0% |
| 4 minus N squared |
| squared plus 4n squared |
|
|
| align:start position:0% |
| |
| |
|
|
| align:start position:0% |
| |
| quantity squared square rooted |
|
|
| align:start position:0% |
| quantity squared square rooted |
| |
|
|
| align:start position:0% |
| quantity squared square rooted |
| times sine of n t minus Phi n |
|
|
| align:start position:0% |
| times sine of n t minus Phi n |
| |
|
|
| align:start position:0% |
| times sine of n t minus Phi n |
| where Phi n was given using the |
|
|
| align:start position:0% |
| where Phi n was given using the |
| |
|
|
| align:start position:0% |
| where Phi n was given using the |
| arctangent formula and |
|
|
| align:start position:0% |
| arctangent formula and |
| |
|
|
| align:start position:0% |
| arctangent formula and |
| this gives us a solution which note is |
|
|
| align:start position:0% |
| this gives us a solution which note is |
| |
|
|
| align:start position:0% |
| this gives us a solution which note is |
| periodic with period 2 t.o |
|
|
| align:start position:0% |
| periodic with period 2 t.o |
| |
|
|
| align:start position:0% |
| periodic with period 2 t.o |
| right I i should also note that 5n is |
|
|
| align:start position:0% |
| right I i should also note that 5n is |
| |
|
|
| align:start position:0% |
| right I i should also note that 5n is |
| between |
|
|
| align:start position:0% |
| |
| |
|
|
| align:start position:0% |
| |
| 0 and 2 pi |
|
|
| align:start position:0% |
| 0 and 2 pi |
| |
|
|
| align:start position:0% |
| 0 and 2 pi |
| sorry 0 it and pie |
|
|
| align:start position:0% |
| sorry 0 it and pie |
| |
|
|
| align:start position:0% |
| sorry 0 it and pie |
| and |
|
|
| align:start position:0% |
| and |
| |
|
|
| align:start position:0% |
| and |
| this gives us this one solution to the |
|
|
| align:start position:0% |
| this gives us this one solution to the |
| |
|
|
| align:start position:0% |
| this gives us this one solution to the |
| differential equation with a forcing of |
|
|
| align:start position:0% |
| differential equation with a forcing of |
| |
|
|
| align:start position:0% |
| differential equation with a forcing of |
| sine NT on the right hand side |
|
|
| align:start position:0% |
| sine NT on the right hand side |
| |
|
|
| align:start position:0% |
| sine NT on the right hand side |
| so now what we want to do is we want to |
|
|
| align:start position:0% |
| so now what we want to do is we want to |
| |
|
|
| align:start position:0% |
| so now what we want to do is we want to |
| sum up many of these solutions using the |
|
|
| align:start position:0% |
| sum up many of these solutions using the |
| |
|
|
| align:start position:0% |
| sum up many of these solutions using the |
| superposition principle so if I go back |
|
|
| align:start position:0% |
| superposition principle so if I go back |
| |
|
|
| align:start position:0% |
| superposition principle so if I go back |
| now |
|
|
| align:start position:0% |
| now |
| |
|
|
| align:start position:0% |
| now |
| i'm going to write the original |
|
|
| align:start position:0% |
| i'm going to write the original |
| |
|
|
| align:start position:0% |
| i'm going to write the original |
| differential equation |
|
|
| align:start position:0% |
| |
| |
|
|
| align:start position:0% |
| |
| so i'm just going to go back and just |
|
|
| align:start position:0% |
| so i'm just going to go back and just |
| |
|
|
| align:start position:0% |
| so i'm just going to go back and just |
| rewrite this |
|
|
| align:start position:0% |
| rewrite this |
| |
|
|
| align:start position:0% |
| rewrite this |
| and i'm going to write the right-hand |
|
|
| align:start position:0% |
| and i'm going to write the right-hand |
| |
|
|
| align:start position:0% |
| and i'm going to write the right-hand |
| side using its fourier series 1 over n |
|
|
| align:start position:0% |
| side using its fourier series 1 over n |
| |
|
|
| align:start position:0% |
| side using its fourier series 1 over n |
| sine NT |
|
|
| align:start position:0% |
| sine NT |
| |
|
|
| align:start position:0% |
| sine NT |
| where n is odd |
|
|
| align:start position:0% |
| where n is odd |
| |
|
|
| align:start position:0% |
| where n is odd |
| ok and this is the problem we originally |
|
|
| align:start position:0% |
| ok and this is the problem we originally |
| |
|
|
| align:start position:0% |
| ok and this is the problem we originally |
| wanted to solve and essentially what |
|
|
| align:start position:0% |
| wanted to solve and essentially what |
| |
|
|
| align:start position:0% |
| wanted to solve and essentially what |
| we've done is we've solved the problem |
|
|
| align:start position:0% |
| we've done is we've solved the problem |
| |
|
|
| align:start position:0% |
| we've done is we've solved the problem |
| for each individual sine NT so how do we |
|
|
| align:start position:0% |
| for each individual sine NT so how do we |
| |
|
|
| align:start position:0% |
| for each individual sine NT so how do we |
| get the full solution well what we have |
|
|
| align:start position:0% |
| get the full solution well what we have |
| |
|
|
| align:start position:0% |
| get the full solution well what we have |
| to do is we have to multiply the |
|
|
| align:start position:0% |
| to do is we have to multiply the |
| |
|
|
| align:start position:0% |
| to do is we have to multiply the |
| solution for each sign in T by a factor |
|
|
| align:start position:0% |
| solution for each sign in T by a factor |
| |
|
|
| align:start position:0% |
| solution for each sign in T by a factor |
| of 4 divided by PI and 1 over N and then |
|
|
| align:start position:0% |
| of 4 divided by PI and 1 over N and then |
| |
|
|
| align:start position:0% |
| of 4 divided by PI and 1 over N and then |
| we have to add all of these solutions up |
|
|
| align:start position:0% |
| we have to add all of these solutions up |
| |
|
|
| align:start position:0% |
| we have to add all of these solutions up |
| for all odd values of n |
|
|
| align:start position:0% |
| |
| |
|
|
| align:start position:0% |
| |
| so for example X is just going to be |
|
|
| align:start position:0% |
| so for example X is just going to be |
| |
|
|
| align:start position:0% |
| so for example X is just going to be |
| four over pi |
|
|
| align:start position:0% |
| four over pi |
| |
|
|
| align:start position:0% |
| four over pi |
| sum of n odd |
|
|
| align:start position:0% |
| sum of n odd |
| |
|
|
| align:start position:0% |
| sum of n odd |
| one over n times the solution to every |
|
|
| align:start position:0% |
| one over n times the solution to every |
| |
|
|
| align:start position:0% |
| one over n times the solution to every |
| sine NT |
|
|
| align:start position:0% |
| sine NT |
| |
|
|
| align:start position:0% |
| sine NT |
| which we've already computed and I've |
|
|
| align:start position:0% |
| which we've already computed and I've |
| |
|
|
| align:start position:0% |
| which we've already computed and I've |
| denoted as X sub N and |
|
|
| align:start position:0% |
| denoted as X sub N and |
| |
|
|
| align:start position:0% |
| denoted as X sub N and |
| X sub n is up here |
|
|
| align:start position:0% |
| X sub n is up here |
| |
|
|
| align:start position:0% |
| X sub n is up here |
| so if we want to be explicit about it |
|
|
| align:start position:0% |
| so if we want to be explicit about it |
| |
|
|
| align:start position:0% |
| so if we want to be explicit about it |
| I'll write the whole thing out as |
|
|
| align:start position:0% |
| I'll write the whole thing out as |
| |
|
|
| align:start position:0% |
| I'll write the whole thing out as |
| for over PI and odd |
|
|
| align:start position:0% |
| |
| |
|
|
| align:start position:0% |
| |
| one over |
|
|
| align:start position:0% |
| one over |
| |
|
|
| align:start position:0% |
| one over |
| four minus N squared quantity squared |
|
|
| align:start position:0% |
| four minus N squared quantity squared |
| |
|
|
| align:start position:0% |
| four minus N squared quantity squared |
| plus 4n squared |
|
|
| align:start position:0% |
| plus 4n squared |
| |
|
|
| align:start position:0% |
| plus 4n squared |
| square rooted |
|
|
| align:start position:0% |
| square rooted |
| |
|
|
| align:start position:0% |
| square rooted |
| we also have a factor of n out front |
|
|
| align:start position:0% |
| we also have a factor of n out front |
| |
|
|
| align:start position:0% |
| we also have a factor of n out front |
| and we have a sine NT |
|
|
| align:start position:0% |
| and we have a sine NT |
| |
|
|
| align:start position:0% |
| and we have a sine NT |
| minus Phi n |
|
|
| align:start position:0% |
| |
| |
|
|
| align:start position:0% |
| |
| okay so there's the final answer for a |
|
|
| align:start position:0% |
| okay so there's the final answer for a |
| |
|
|
| align:start position:0% |
| okay so there's the final answer for a |
| particular solution in its full glorious |
|
|
| align:start position:0% |
| particular solution in its full glorious |
| |
|
|
| align:start position:0% |
| particular solution in its full glorious |
| detail we can also check that this |
|
|
| align:start position:0% |
| detail we can also check that this |
| |
|
|
| align:start position:0% |
| detail we can also check that this |
| particular solution is periodic note how |
|
|
| align:start position:0% |
| particular solution is periodic note how |
| |
|
|
| align:start position:0% |
| particular solution is periodic note how |
| each sign in T is periodic and each sine |
|
|
| align:start position:0% |
| each sign in T is periodic and each sine |
| |
|
|
| align:start position:0% |
| each sign in T is periodic and each sine |
| NT has period of at least two pi so when |
|
|
| align:start position:0% |
| NT has period of at least two pi so when |
| |
|
|
| align:start position:0% |
| NT has period of at least two pi so when |
| we sum up a whole bunch of functions |
|
|
| align:start position:0% |
| we sum up a whole bunch of functions |
| |
|
|
| align:start position:0% |
| we sum up a whole bunch of functions |
| which all have a period of at least two |
|
|
| align:start position:0% |
| which all have a period of at least two |
| |
|
|
| align:start position:0% |
| which all have a period of at least two |
| pi the the sum is also going to be a |
|
|
| align:start position:0% |
| pi the the sum is also going to be a |
| |
|
|
| align:start position:0% |
| pi the the sum is also going to be a |
| periodic function with at least two pi |
|
|
| align:start position:0% |
| periodic function with at least two pi |
| |
|
|
| align:start position:0% |
| periodic function with at least two pi |
| and |
|
|
| align:start position:0% |
| and |
| |
|
|
| align:start position:0% |
| and |
| and as a result this gives us that the |
|
|
| align:start position:0% |
| and as a result this gives us that the |
| |
|
|
| align:start position:0% |
| and as a result this gives us that the |
| the answer we're looking for also in |
|
|
| align:start position:0% |
| the answer we're looking for also in |
| |
|
|
| align:start position:0% |
| the answer we're looking for also in |
| addition if we want the full general |
|
|
| align:start position:0% |
| addition if we want the full general |
| |
|
|
| align:start position:0% |
| addition if we want the full general |
| solution to the differential equation to |
|
|
| align:start position:0% |
| solution to the differential equation to |
| |
|
|
| align:start position:0% |
| solution to the differential equation to |
| this particular solution we also have to |
|
|
| align:start position:0% |
| this particular solution we also have to |
| |
|
|
| align:start position:0% |
| this particular solution we also have to |
| add the homogeneous piece |
|
|
| align:start position:0% |
| add the homogeneous piece |
| |
|
|
| align:start position:0% |
| add the homogeneous piece |
| so this concludes the problem and I'll |
|
|
| align:start position:0% |
| so this concludes the problem and I'll |
| |
|
|
| align:start position:0% |
| so this concludes the problem and I'll |
| just quickly recap when solving a |
|
|
| align:start position:0% |
| just quickly recap when solving a |
| |
|
|
| align:start position:0% |
| just quickly recap when solving a |
| differential equation with the periodic |
|
|
| align:start position:0% |
| differential equation with the periodic |
| |
|
|
| align:start position:0% |
| differential equation with the periodic |
| forcing function on the right hand side |
|
|
| align:start position:0% |
| forcing function on the right hand side |
| |
|
|
| align:start position:0% |
| forcing function on the right hand side |
| again to iterate the steps you first |
|
|
| align:start position:0% |
| again to iterate the steps you first |
| |
|
|
| align:start position:0% |
| again to iterate the steps you first |
| before you decompose the right hand side |
|
|
| align:start position:0% |
| before you decompose the right hand side |
| |
|
|
| align:start position:0% |
| before you decompose the right hand side |
| into summation of sines and cosines you |
|
|
| align:start position:0% |
| into summation of sines and cosines you |
| |
|
|
| align:start position:0% |
| into summation of sines and cosines you |
| then solve the differential equation for |
|
|
| align:start position:0% |
| then solve the differential equation for |
| |
|
|
| align:start position:0% |
| then solve the differential equation for |
| sine NT cosine NT |
|
|
| align:start position:0% |
| sine NT cosine NT |
| |
|
|
| align:start position:0% |
| sine NT cosine NT |
| individually this gives you a solution |
|
|
| align:start position:0% |
| individually this gives you a solution |
| |
|
|
| align:start position:0% |
| individually this gives you a solution |
| for each term on the right hand side of |
|
|
| align:start position:0% |
| for each term on the right hand side of |
| |
|
|
| align:start position:0% |
| for each term on the right hand side of |
| the differential equation and then at |
|
|
| align:start position:0% |
| the differential equation and then at |
| |
|
|
| align:start position:0% |
| the differential equation and then at |
| the end of the day you superposition to |
|
|
| align:start position:0% |
| the end of the day you superposition to |
| |
|
|
| align:start position:0% |
| the end of the day you superposition to |
| sum up all this all of the solutions and |
|
|
| align:start position:0% |
| sum up all this all of the solutions and |
| |
|
|
| align:start position:0% |
| sum up all this all of the solutions and |
| that gives you one final big solution |
|
|
| align:start position:0% |
| that gives you one final big solution |
| |
|
|
| align:start position:0% |
| that gives you one final big solution |
| okay so i hope you enjoyed this problem |
|
|
| align:start position:0% |
| okay so i hope you enjoyed this problem |
| |
|
|
| align:start position:0% |
| okay so i hope you enjoyed this problem |
| and i'll see you next time |
