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This year I'm having trouble with my Signals and Systems class. My major subject is Software Engineering and Electric and Electrical Engineering is my Minor. This question was my previous exam question but I coudn't write anything about it as a solution. If somebody could at least show a way to solve this problem it would be great.

Question:

$$\frac{d^2y(t)}{d^2t^2}+\frac{6dy(t)}{d(t)}+8y(t)=2x(t)$$ a) Find impulse response $h(t)$
b) Find output $y(t)$ if $x(t)=te^{-2t}u(t)$

I know that $h(t) = y(t)/x(t)$, and i m guessing that i can get that using z-transform but I don't know how to and if it is the correct way.

Thanks in advance...

EDIT:
Here is what I have done so far. Photo of my Work Done

From here, to find the output I think I will use convolution.? y(t) = h(t)*x(t)

I used Laplace transform to find the inverse fourier transform of the function H(jw). What was I going to do if Laplace transform would not be suitable to situation?

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  • $\begingroup$ My comment to this quesiton also applies to you. $\endgroup$ – Matt L. Jan 5 '16 at 8:08
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    $\begingroup$ Hint: Laplace Transform. $\endgroup$ – Aenid Jan 5 '16 at 9:04
  • $\begingroup$ I made a mistake at my first upload, I have corrected it and uploaded again. Not sure if it is correct or not but this is all i can come up with so far. $\endgroup$ – M. Jay Jan 5 '16 at 12:39
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    $\begingroup$ Off Topic: Consider reversing your minor and major... :) $\endgroup$ – Fat32 Jan 5 '16 at 13:32
  • $\begingroup$ I didn't understand why you said that ? Was is because of the orientation of the picture or the solution ? I didn't like signals class so maybe I will quit from minor program. @Fat32 $\endgroup$ – M. Jay Jan 5 '16 at 18:06
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It looks like your transfer function is correct, but there's a small mistake in your partial fraction expansion:

$$H(s)=\frac{2}{(s+4)(s+2)}=\frac{1}{s+2}-\frac{1}{s+4}\tag{1}$$

The corresponding impulse response is

$$h(t)=(e^{-2t}-e^{-4t})u(t)\tag{2}$$

The response to $x(t)=te^{-2t}u(t)$ is indeed most easily computed by solving the convolution integral:

$$y(t)=u(t)\int_0^tx(\tau)h(t-\tau)d\tau\tag{3}$$

I leave the exercise of solving $(3)$ up to you, but if I'm not mistaken the result should be

$$y(t)=\frac{1}{4}e^{-2t}\left[2t^2-2t+1-e^{-2t}\right]u(t)\tag{4}$$

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  • $\begingroup$ You are right! I wrote 2/2 = 2. That was a silly mistake. Now I started to understand things. I'm not good at convolution so I will try multiplying h(jw) and x(jw) to obtain the output. If I can't achive to result that way, I will give a try to convoluton. Thank you ! @Matt $\endgroup$ – M. Jay Jan 6 '16 at 1:27
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Let me provide a method, for part a, applied only for finding the impulse response $h(t)$ of an LTI system characterised by an LCCDE of the form $ \sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = \sum_{k=0}^{M}{ b_k {{d^k x(t)}\over {dt^k}}}$ by using the classical time domain approach. A method which is generally ignored.

Note that the input $x(t) = \delta(t)$ is formally a problematic signal (function) and therefore mathematically oriented classical books on ODEs, do generally avoid any discussion of such functions and their solutions, unless the scope of the book specifically includes generalised functions (distributions), usually to be used in some engineering or physical fields.

The procedure: Consider an LTI system which is causal with initial rest conditions. We say that when the input to this system is $x(t)=\delta(t)$ an impulse, then its output $y(t)=h(t)$ is the impulse-response of the system. We find this solution $h(t)$ in two steps, by breaking the LCCDE into two parts as inspired by a serial (cascade) implementation of two LTI systems corresponding to the right and left sides of the LCCDE.

I/O relationship of Part-I is given by the LCCDE: $$\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = x(t)$$ which represents the first stage with its input $x(t)=\delta(t)$ and we denote its solution (stage1-output) as $h_0(t)$ which is actually the impulse response of Part-I.

And the I/O relationship of the Part-II is given by the equation: $$ y(t) = \sum_{k=0}^{M}{b_k{{d^k x(t)}\over {dt^k}}} $$ which requires nothing but simple summation of its input, $x(t) = h_0(t)$ ,and its derivatives to compute the output as $$h(t) = \sum_{k=0}^{M}{b_k {{d^k h_0(t)}\over {dt^k}}}$$ Therefore we need to find $h_0(t)$ of the Part-I to simply compute the impulse response $h(t)$ of the complete system.

To find the solution of part-I: $\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = x(t)$ whose solution $y(t)$ is what we call $h_0(t)$ , we we make the following observation that $y(t)$ is composed of two parts as $$y(t) = y_{h}(t) + y_{p}(t)$$ where $y_{h}(t)$ is the homogeneous solution corresponding to $x(t)=0$ for all t, and $y_{p}(t)$ is the particular solution corresponding to input $x(t)$ for $ t > 0$, in particular as what remains as input when time goes to infinity.

Now, we can see that for the input $x(t)=\delta(t)$ the particular solution $y_p(t)$ for $t>0$ is zero as the input is zero for $t > 0$. Then the output simplifies to $h_0(t) = y(t) = y_h(t)$

Now, we have to find the homogeneous part of $y(t)$ as the solution of the equation $$\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = 0$$ following the usual procedure by finding the roots of its characteristic polynomial in complex-s: $$p(s) = \sum_{k=0}^{N}{a_k s^k}$$

which is obtained when we simply insert $y_h(t) = Ke^{st}$ as a solution canditade to the homogeneous differential equation $\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = 0$. The solution is then hypothesised to be $$y_h(t) = \sum_{k=0}^{N}{ A_k e^{st}}$$ for the case when all the roots are distinct. If they are not, you know what to do right ;) The solution therefore now requires to determine the N unknown coefficients $A_k$ from N-inital conditions, at time t=0 for the framework of this method, possesed by the output $y(t)$ and its derivates up to order N-1.

Remember that this system was defined to be under inital rest with complete zero inital conditions prior to the application of any excitation. Hence if we were to solve the homogeneous solution $y_h(t)$ for an ordinary (formally valid) input signal $x(t)$ to the system, the homogeneous solution will be identically zero, as the initial conditions would all be. (all the coefficients $A_k$ of the exponential terms turn out to be zero). But the magic of the generalised input $\delta(t)$ is that it will set one of those inital conditions to non-zero, thereby, enabling a non-zero homogeneous response to exist, which will become the solution of the part-I as well.

It can be shown that the inital conditions of the part-I due to the excitation $\delta(t)$ will be set as $y(t)=0 ~,~ y(t)' = 0 ~,~ y(t)''=0 ~,~... ~,~ y(t)^{N-1}=1/a_N$ all at $t=0$.

From this information you can proceed to solve the algebraic equaitons as usual and find $h_0(t)$ from which you can compute the impulse response $h(t)$ for all $t > 0$. What happens for all $t <0 $? As we stated, this system was LTI and causal with initial rest condition. For such a system if the input $x(t)$ is zero for all $t<0$ then that output is also zero for all $t<0$ which is achieved by multiplying $h_0(t)$ via $u(t)$ to represent the solution of Part-I for all t.

It is instructive to apply this procedure to your example: $$y'' + 6 y' + 8y = 2x$$ Lets first solve Part-I: $$y'' + 6 y' + 8y = x$$ to find $h_0(t)$ when the input $x(t)$ = $\delta(t)$. solve the characteristic equation: $s^2 + 6s + 8 = 0$ which yields $s=-4$ and $s=-2$ as solutions. A stable system, isn't it? The proposed homogeneous solution is therefore $$y_h(t) = A_1 e^{-2t} + A_2 e^{-4t}$$ For which we need 2 inital conditions to find $A_1$ and $A_2$. These initial conditions are set by the impulse $\delta(t)$ to be: $y(0)=0$ and $y(0)'=1$. Using these we find them to be $A_1=1/2$ and $A_2=-1/2$ and this yields $$ h_0(t) = {1 \over 2} [ e^{-2t} - e^{-4t}] $$ for all $t > 0$. Also note that this system was LTI and causal with initial rest and input $x(t)=\delta(t)$ is zero for all $t<0$ implies that output is also zero for all $t<0$ which is simplified by using a unit step $u(t)$ function to format the output as: $$ h_0(t) = {1 \over 2} [ e^{-2t} - e^{-4t}]u(t)$$ for all t.

Finally apply this $h_0(t)$ as an input to Part-II whose output is simply $$y(t)= 2x(t)$$ to get the impulse response of the overall system as: $$h(t) = [ e^{-2t} - e^{-4t}]u(t)$$

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  • $\begingroup$ Fat32, how do you apply your method to the following Differential Equation?: y''+3y'+2y=x''+x'-2x I tried your method but I seem to get a wrong answer... $\endgroup$ – Robert Dec 7 at 15:42
  • $\begingroup$ @Robert I've checked the procedure and it works as expected. (At first I thought that the pole-zero cancellation could be a problem, but no... Applying the procedure yields the correct answer.) Just be sure to take the initial condition for $y(0)'$ as $1/a_N = 1/1$ where $a_N=1$ is the coefficient of the highest derivative of $y(t)$. $\endgroup$ – Fat32 Dec 9 at 12:36

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