Question

Question 13 : Signals:

Consider the system described by :

\(y^{\prime} (t) + 2y(t) = x(t) + x^{\prime}(t)\)

Find the \(\color{green}{\textsf{Impulse Response h(t)}}\) of the system..

How to solve this Differential Equation for both Homogeneous as well as Particular Solution??

Answer 1

If \(x(t) = \delta(t)\), then we will get: \(y(t) = h(t)\) which we have to find..

So, we have got:

\[h(t) + 2h^{\prime}(t) = \delta(t) + \delta^{\prime}(t)\]

Answer 2

Here, we have to find the solution for \(h(t)\) both Particular and Homogeneous..

Homogeneous will be when:

\[h(t) + 2h^{\prime}(t) = 0\]

Answer 3

@Miracrown

Answer 4

@UnkleRhaukus , I am missing you badly here.. :P

Answer 5

\[\begin{align}
h(t)+2h'(t)&=0\\
%2h'(t)&=-h(t)\\
%\frac{2}{-h(t)}h'(t)&=0\\
\frac{2}{-h(t)}\frac{\mathrm dh}{\mathrm dt}&=0\\
\frac{\mathrm dh}{h(t)}&=0\\
\int\frac{\mathrm dh}{h(t)}&=0\\
\ln|h(t)|+c &=0\\
%\ln|h(t)| &=-c\\
h(t) &= e^{-c}\\
\\
h(t)&=C
\end{align}\]

Answer 6

How?? I am not good at it, you should also tell how you did it.. :(

Answer 7

How did you get second step??

Answer 8

\[h(t) + 2h^{\prime}(t) = 0 \implies 2 h^{\prime}(t) = -h(t)\]

\[\frac{2h^{\prime}(t)}{-h(t)} = 1\]

How you got 0 there?

Answer 9

***

\[\begin{align}
h(t)+2h'(t)&=0\\
2h'(t)&=-h(t)\\
%\frac{2}{-h(t)}h'(t)&=1\\
%\frac{2}{-h(t)}\frac{\mathrm dh}{\mathrm dt}&=1\\
\frac{\mathrm dh}{h(t)}&=-\frac{\mathrm dt}2\\
\int\frac{\mathrm dh}{h(t)}&=-\int\frac{\mathrm dt}2\\
\ln|h(t)|+c_1 &=-\frac t2+c_2\\
&& c = c_2-c_1\\
\ln|h(t)| &=c-\frac t2\\
h(t) &= e^{c-t/2}\\
h(t) &= e^{c}e^{-t/2}\\
&&C = e^c\\
\\
h(t)&=Ce^{-t/2}
\end{align}\]

Answer 10

sorry about previous mistake

Answer 11

Still my book says : It should be : \(h(t) = c \cdot e^{-2t}\) ..

Answer 12

We can use Linear Method to solve it:

We know for:
\[\frac{dy}{dt} + P = Q \quad \quad ; \text{P and Q are functions of t} \\ \text{Integrating Factor is : } e^{\int\limits{P.dt}}\]

And Solution is given by:

\[y \cdot (IF) = \int\limits (Q \cdot (IF)) \cdot dt + c\]

Answer 13

This is the Particular solution, if we set Q = 0, and then find solution, then we will get Homogeneous Solution to this DE.. Are you getting what I am trying to say??

Answer 14

Oh, \(y\) is missing there..

\[\frac{dy}{dt} + Py = Q\]

Answer 15

for the homogenous equation, I find the "characteristic equation"
y' + 2y = 0
m + 2 = 0
and m= -2
the solution will be exp(m t), so
y_h= c1 e^(-2t)

Answer 16

Yes, it is easier one..

Answer 17

But how to go for Particular Solution?

Answer 18

Be remember Input is \(\delta(t)\) , A Dirac-Delta Function, which is undefined for t = 0, and which is 0 for any other t..

Answer 19

\[h^{\prime}(t) + 2h(t) = \delta(t) + \delta^{\prime}(t)\]

Answer 20

where \(h(t)\) is the Impulse Response when \(\delta(t)\) is applied as Input..

Answer 21

the derivative of the impulse is the unit step, call it the heaviside function H(t)
y' + 2y = D(t) + H(t)

let y= C H(t). Plug into the equation
C D(t) + 2C H(t) = D(t) + H(t)
I think it's safe to say C D(t) = D(t) (D(t) is "infinity" at t=0, 0 otherwise)
so C= ½
and we get
y_p= H(t)/2

y= c exp(-2t) + H(t)/2

that is close to wolfram's answer.
http://www.wolframalpha.com/input/?i=solve+y%27%2B2y%3D%3DDiracDelta%5Bt%5D+%2B+HeavisideTheta%5Bt%5D

but I must be missing some details

Answer 22

The derivative of Impulse is not Unit Step Function..

Answer 23

The Integral of Impulse is an Unit Step Function..

Or in other words, the derivative of Unit Step is Impulse..

Answer 24

I thought I was doing derivative of the unit step?

Answer 25

My book says: Particular Solution is : \(h(t) = c_1 \delta(t)\).. This is what they have assumed..

Answer 26

I want to know that how you initially assumes a solution to be??

How I can say let the solution be so and so, then by plugging in the solution in the equation, I find constants what I have assumed in my solution.. I mean how you can analyze this must be solution to this particular DE??

Answer 27

Homogeneous solution is okay, we are right in that but how they can initially assume that let the solution be in terms of \(\delta(t)\) ??

Answer 28

Anyways the book I am following is :

\[\color{green}{\textsf{Schaum's Outlines Of Signals and Systems..}}\]

Answer 29

This looks like a nice way to solve the problem, using fourier transforms and their properties. using the tricks of ODE seems difficult (because the delta and unit-step are not really "ordinary".

Answer 30

Laplace is yet more simpler.. :) You don't have to write two characters always ie \(j \omega\), write simply \(s\)..

Answer 31

But, yes Fourier Transform and Laplace Transforms are yet easier method to go with, but I am learning these for my knowledge only..

Answer 32

I will now move on to Fourier Series, and then Transform, so there is still a time to use Fourier Transforms and Laplace here.. :)

Answer 33

May be @unkleRhaukus will want to explain it more to me.. :)

Answer 34

maybe, but I think the methods used to solve for the particular solution for this type of problem probably all rely on transforms. in other words, the method of "undetermined coefficients" or "variation of parameters" are not applicable (those being the only the methods I have seen for solving for the particular solution)

Answer 35

Or may be: It is just by analyzing, or may be saying like : that we have got derivative of delta function on the right hand side, so the particular solution will contain delta as one of its terms..

Answer 36

@hartnn can you just see this?