# How do I find the transfer function in the frequency domain?

I was doing some exercises with transfer functions, they were always under the form of $$H(z)$$ and $$H(e^{jw})$$ for the frequency response. Today I have found one with $$H(f)$$. I would like to ask if my solution to the problem is good ? Let $$y(n) = -x(n)$$ where $$y(n)$$ is the output and $$x(n)$$ is the input. Find the transfer function $$H(f)$$.

Here is my approach to the problem:

$$y(n) = -x(n) => H(n) = \frac{y(n)}{x(n)} = -1$$

We apply the Fourier Transform

$$H(f) = F\{H\} = F\{-1\}= \int_{-\infty}^{\infty}e^{j2\pi f t}dt = -\delta(f)$$

Since $$\delta(f) = 1$$ we conclude that $$H(f) = -1$$

Your final answer looks good, but the derivation and notation are a bit mixed. Remember the capitalized $$H()$$ refers to the frequency domain transfer function, so writing $$H(n)$$ with the time-domain index $$n$$ doesn't make sense. Also, it is the time-domain impulse response that is a delta function, not the frequency domain transfer function, so $$H(f)=-\delta(f)$$ is not right.

The problem can be solved very easily:

$$y(n)=-x(n)$$

Take the Fourier Transform of both sides:

$$Y(f)=-X(f)$$

Divide both sides by $$X(f)$$:

$$\frac{Y(f)}{X(f)} = H(f) = -1$$

In the time domain:

$$h(n)=-\delta(n)$$