# FIR Impulse response & transfer function

I am working though a dsp past paper and I have come across the questions "find the impulse response in the time domain" & "find the transfer function in the time domain" I know its really simple but the "time domain" part of the question catches me out, I know how to find the transfer function in the Z domain, its not for homework, its a past paper I am looking through.

How would I find the impulse response in the time domain?

AND

How would I find the transfer function in the time domain?

I have already tried YouTube and google and a few books but haven't found anything that is specific to those questions. The question is attached in the photo.

• For the impulse response you simply apply an impulse at the input and see what you get at the output. "Time domain transfer function" does not make any sense. The transfer function is just the Z-transform of the impulse response. – Matt L. Jul 15 '18 at 7:49
• @MattL., it's poor semantics, but "time domain transfer function" might mean the input-output difference equation. it can make some sense in that it maps the time-domain input to the time-domain output, it's just that you don't multiply with it. – robert bristow-johnson Jul 16 '18 at 23:54

I agree with Matt's comment, however i suspect that whoever gave you the assignment wants you to write down an equation describing the transfer behaviour in time domain for an arbitrary input.

When $x[n] = \delta[n]$, the impulse response of the given system would be

$$h[n] = -\delta[n] + 2\delta[n-1] - 2\delta[n-2] \ .$$

You can find it in the output $y[n]$ by setting the input $x[n]$ to

$$x[n] = \delta[n] \triangleq \begin{cases} 1 \qquad \text{if } n = 0 \\ 0 \qquad \text{if } n \ne 0 \\ \end{cases}$$

Then the output at the end of the diagram is $h[0]=-1$. The $z^{-1}$ elements act as delays of one sample, so the next outputs are $h[1]=2$ and $h[2]=-2$, respectively.

Now to find out what the equation does to general inputs in time domain, write the same thing down in terms of $x[n]$ instead of $x[n]=\delta[n]$.

For any given $x[n]$, the resulting $y[n]$ is produced by the current input times $-1$ and the prior two inputs multiplied with their factors $2$ and $-2$. You can then write down the general solution as

$$y[n] = -x[n] + 2x[n-1] -2x[n-2] \ .$$

Keep in mind however that the term "time domain transfer function" could be misunderstood by some, which is probably the reason your search did not yield any results.