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(c)A linear time-invariant system with impulse response $h(t)=e^{-\alpha t} u(t)$ is driven by the input $x(t)=e^{-\beta t} u(t)$ . It is assumed that $\alpha, \beta > 0$.

  • (i) Using frequency domain analysis, find the output of the system. [Hint: $Y(f) = X(f) H(f)$]

I'm finding in trouble trying to resolve this exercise. I have to calculate the convolution of this signal:

well I applied the formula that says that the convolution of this two signal as state by the hint, equal to

$Y(f)=X(f)H(f)$ where $H(f)$ is the fourier transform of the impulse response and $X(f)$ is the fourier transform of input signal.

well fourier transform of $h(t)=e^{−bt}u(t)$ is $H(f)=1/(b+j2πf)$. fourier transform of input signal $x(t)=e^{−at}u(t)$ is $X(f)=1/(a+j2πf)$

Thus,

$Y(f)=X(f)H(f)$

$$Y(f)=\frac{1}{a+j2πf}\cdot \frac{1}{b+j2πf}$$

is this a correct step ?

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    $\begingroup$ This looks like some homework. What techniques did you already try? What is your specific problem? $\endgroup$ – Maximilian Matthé Feb 14 '17 at 8:34
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    $\begingroup$ For this kind of problem (homework-style), you normally show what you yourself have done, your effort into it, then tell us where you're stuck. Then people here can see how to give you some hints so you can understand the problem and eventually find the solution. $\endgroup$ – Gilles Feb 14 '17 at 8:34
  • $\begingroup$ Such a dreadful problem to suggest for frequency-domain analysis when time-domain analysis is so much simpler. It is truly said: when all you have is a hammer, everything looks like a nail. $\endgroup$ – Dilip Sarwate Feb 15 '17 at 22:56
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It is indeed correct to say that

$$Y(j\omega)=\frac{1}{\alpha+j\omega}\cdot\frac{1}{\beta+j\omega}=\frac{1}{(\alpha+j\omega)(\beta+j\omega)}$$

You could write the answer just in that why. The problem with that is that it would be a bit hard to antitransform and come back to the time-domain to find $y(t)$. So let's manipulate that expression a little bit. Using the partial-fraction expansion we get:

$$Y(j\omega)=\frac{1}{(\alpha+j\omega)(\beta+j\omega)}=\frac{A}{\alpha+j\omega}+\frac{B}{\beta+j\omega}$$

where we want to calculate the constants $A$ and $B$. Rewriting those two fractions as one we get that:

$$\frac{A}{\alpha+j\omega}+\frac{B}{\beta+j\omega}=\frac{A(\beta+j\omega)+B(\alpha+j\omega)}{(\alpha+j\omega)(\beta+j\omega)}=\frac{A\beta+jA\omega+B\alpha+jB\omega}{(\alpha+j\omega)(\beta+j\omega)}$$

So:

$$\frac{1}{(\alpha+j\omega)(\beta+j\omega)}=\frac{A\beta+jA\omega+B\alpha+jB\omega}{(\alpha+j\omega)(\beta+j\omega)}$$

That equality holds only if

$$1=A\beta+jA\omega+B\alpha+jB\omega=(A\beta+B\alpha)+j\omega(A+B)$$

As the left side of the equation is purely real, then the imaginary part of the right side must equal zero and we can calculate our constants as functions of $\alpha$ and $\beta$:

$$A=-B \implies 1 = A\beta - A\alpha \implies A =\frac{1}{\beta-\alpha} \implies B=-\frac{1}{\beta-\alpha}$$

So we can write the transform of the output as proposed in the beginning:

$$Y(j\omega)= \frac{A}{\alpha+j\omega}+\frac{B}{\beta+j\omega}= \frac{\frac{1}{\beta-\alpha}}{\alpha+j\omega}-\frac{\frac{1}{\beta-\alpha}}{\beta+j\omega}$$

Therefore our output is:

$$Y(j\omega) = \frac{1}{\beta-\alpha}\left(\frac{1}{\alpha+j\omega}-\frac{1}{\beta+j\omega}\right)$$

And now you only have to antitransform that (I believe you can do that) and that's your output in the time domain.

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