# fourier / convolution

#### hamza324

Joined Jul 10, 2011
33
HI
My prof gave a question

$$x= \delta(t-1) + 5Cos(2t) \ \ \ \longrightarrow \ \ \ \ \ \fbox{H( \omega)= \frac {1}{j\omega +1}} \ \ \ \longrightarrow \ \ \ y=find \ using \ super \ position$$

Now for the solution , he treats each of the input functions separately with the boxed function.

he writes

$$x_1 = \delta(t-1) \longrightarrow \fbox{h(t)=e^{-t} u(t)} \longrightarrow h(t-1)=e^{-(t-1)}u(t-1) = y_1$$

Then,

$$x_2 = 5Cos(2t) \longrightarrow \fbox {H(\omega)=\frac{1}{j\omega+1}} \longrightarrow 5\ \mid H(\omega)\mid \ Cos(2t + \theta)$$

$$where \mid H(\omega) \mid=( \frac{1}{(2)^2+1})^{\frac{1}{2}}and \theta =- tan^{-1} ( \frac{2}{1})=-63^o$$

So using super sposition , we get

$$y(t)=\ e^{-(t-1)} u(t-1)\ \ \ + \ \ \ \frac{5}{\sqrt{5}} Cos(2t-63^o)$$

Now my question is that,

1 - Why did he used time domain for the impulse function and frequency domain for cos(2t).

2 - How did he get expression for $$y_2$$ , the one with magnitude of H(w).

thnx

#### steveb

Joined Jul 3, 2008
2,436
HI
My prof gave a question

$$x= \delta(t-1) + 5Cos(2t) \ \ \ \longrightarrow \ \ \ \ \ \fbox{H( \omega)= \frac {1}{j\omega +1}} \ \ \ \longrightarrow \ \ \ y=find \ using \ super \ position$$

Now for the solution , he treats each of the input functions separately with the boxed function.

he writes

$$x_1 = \delta(t-1) \longrightarrow \fbox{h(t)=e^{-t} u(t)} \longrightarrow h(t-1)=e^{-(t-1)}u(t-1) = y_1$$

Then,

$$x_2 = 5Cos(2t) \longrightarrow \fbox {H(\omega)=\frac{1}{j\omega+1}} \longrightarrow 5\ \mid H(\omega)\mid \ Cos(2t + \theta)$$

$$where \mid H(\omega) \mid=( \frac{1}{(2)^2+1})^{\frac{1}{2}}and \theta =- tan^{-1} ( \frac{2}{1})=-63^o$$

So using super sposition , we get

$$y(t)=\ e^{-(t-1)} u(t-1)\ \ \ + \ \ \ \frac{5}{\sqrt{5}} Cos(2t-63^o)$$

Now my question is that,

1 - Why did he used time domain for the impulse function and frequency domain for cos(2t).

2 - How did he get expression for $$y_2$$ , the one with magnitude of H(w).

thnx
For question 1, one is free to use either the frequency domain or them time domain, depending on which is easier. Note that the first input is a delayed impulse function, so the output of that will be a delayed impulse response. So, he transforms the H(w) of the system and then delays the time domain impulse response function.

For question 2, he simply noted that H(w) is the frequency domain response and the input is a pure cosine wave, with frequency 2 and no phase delay. So, he can quickly calculate that the output will be scaled in magnitude and phase shifted by using H(w).

#### hamza324

Joined Jul 10, 2011
33
Thanks for explanation.

I have another similar question regarding discrete time

I have an expression

$$e^{j \Omega n} \longrightarrow \fbox{h[n] \longleftrightarrow H(e^{j \Omega})} \longrightarrow H(e^{j \Omega}) e^{j \Omega n}$$

I know this is frequeny response but i dont understand the expression at the output

Can you please provide me with some simpler explanation for this expression and some kind of example if possible.

#### steveb

Joined Jul 3, 2008
2,436
Thanks for explanation.

I have another similar question regarding discrete time

I have an expression

$$e^{j \Omega n} \longrightarrow \fbox{h[n] \longleftrightarrow H(e^{j \Omega})} \longrightarrow H(e^{j \Omega}) e^{j \Omega n}$$

I know this is frequeny response but i dont understand the expression at the output

Can you please provide me with some simpler explanation for this expression and some kind of example if possible.
This is similar to the analysis of the second component of the input signal above (i.e. the 5cos(2t) part). That example was a continuous time cosine wave going through a system with known frequency response.

This problem is pretty much the same, only it is a complex sine wave, and the problem is in discrete time.

For background, in continuous time the frequency response is obtained by taking the Laplace s-domain transfer function $$H(s)$$, where $$s=\sigma+j\omega$$, and substituting $$s=j\omega$$, which puts you on the jw-axis. In discrete time, the frequency response is obtained by taking the Laplace z-domain transfer function $$H(z)$$, where $$z=r\; {\rm e}^{j\Omega}$$, and substituting $$z={\rm e}^{j\Omega}$$, which puts you on the unit circle.