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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 $H(f) = -1$

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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:

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

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