Calculating impulse response in convolution.

[ace_terabyte]

Homework Statement

y(t) = $$\int ^{t}_{-\infty} e^{-(t-\tau)} x(\tau-2)d\tau$$
Find h(t), given that the continuous time linear time invariant signal has input x(t) and output y(t).

2. The attempt at a solution
I was wondering if I can simply manipulate the expression until i get it in the form y(t) = x(t) * h(t) but I think that that approach is simply wrong. I just have no idea on how to approach this problem and was wondering if you can guide me in the right direction.

 

[CEL]

Homework Statement

y(t) = $$\int ^{t}_{-\infty} e^{-(t-\tau)} x(\tau-2)d\tau$$
Find h(t), given that the continuous time linear time invariant signal has input x(t) and output y(t).

2. The attempt at a solution
I was wondering if I can simply manipulate the expression until i get it in the form y(t) = x(t) * h(t) but I think that that approach is simply wrong. I just have no idea on how to approach this problem and was wondering if you can guide me in the right direction.

Convolution is represented by a product only in the frequency domain: Y(s)=X(s)H(s).
In the time domain convolution is defined by the integral
y(t) = $$\int ^{t}_{-\infty} x(\tau) h(t-\tau)d\tau$$

 

[ace_terabyte]

Thanks for that, the question asks me to find out h(t) and does not give me any other information. I assume it is the same in both frequency and time domain.

 

[CEL]

Thanks for that, the question asks me to find out h(t) and does not give me any other information. I assume it is the same in both frequency and time domain.

NO, in the time domain you have h(t), in the frequency domain you have H(s), the Laplace transform of h(t).
Compare your equation with mine and you can obtain h(t).

 

[ace_terabyte]

y(t) = $$\int ^{t}_{-\infty} e^{-(t-\tau)} x(\tau-2)d\tau$$

So going by what you said:

y(t) = $$\int ^{t}_{-\infty} (\frac{1}{e})^{(t-\tau-2)} x(\tau)d\tau$$

= $$\int ^{t}_{-\infty} (\frac{1}{e})^{(t-\tau)} (\frac{1}{e})^{(2)}x(\tau)d\tau$$

=$$(\frac{1}{e})^{(2)} \int ^{t}_{-\infty} (\frac{1}{e})^{(t-\tau)} x(\tau)d\tau$$

so is h(t) = $$(\frac{1}{e})^{(t-\tau)}$$ ?

 

[CEL]

y(t) = $$\int ^{t}_{-\infty} e^{-(t-\tau)} x(\tau-2)d\tau$$

So going by what you said:

y(t) = $$\int ^{t}_{-\infty} (\frac{1}{e})^{(t-\tau-2)} x(\tau)d\tau$$

= $$\int ^{t}_{-\infty} (\frac{1}{e})^{(t-\tau)} (\frac{1}{e})^{(2)}x(\tau)d\tau$$

=$$(\frac{1}{e})^{(2)} \int ^{t}_{-\infty} (\frac{1}{e})^{(t-\tau)} x(\tau)d\tau$$

so is h(t) = $$(\frac{1}{e})^{(t-\tau)}$$ ?

No, $$h(t-\tau) =(\frac{1}{e})^{(t-\tau)+2}$$

 

[ace_terabyte]

Thanks a lot CEL, I appreciate it.