How does the number of frequency-domain sampling points influence the outcome of an inverse FFT?

Question

Given some frequency-domain representation of an impulse response.

How does the number of frequency-domain sampling points influence the outcome of an inverse FFT, if I keep the sampling frequency constant?

Example 1 Take x(t) = e^(-t)*sin(t) for t in (0,10). Sample the signal, take its fft. Omit every second point of the fft and on the result do an inverse fft. Here is the result:

So it looks as if the frequency-domain downsampling had removed the second portion of my signal, while dividing the first part by two. (Indeed multiplying the result by two results in a curve lying exactly above the first part of the original signal.)

• Why did it remove the second part and not some arbitrary other part? I guess this is related to the second part being nearly zero everywhere, but I can't seem to find the exact reason.
• Why is the result divided by two?
• How does this generalize to other time signals?

Example 2 (pathological) Same as above, this time with a signal that is linearly decreasing from 1 to 0.1.

Now what is that?

Edit There was a mistake here: I accidentally used an uneven number of sampling points, which resulted in a complex signal after the ifft. Here is the fixed plot.

I'm interested in mathematical/theoretical as well as intuitive explanations of what's going on, as well as pointers on where to read on. I even had trouble finding a correct term for what I am doing.

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Here's the R code that produced the above plots.

# First plot
t <- seq(0.01, 10, 0.01)
z <- exp(-t)*sin(t)
Z <- fft(z)/length(z)
Z2 <- Z[seq(1,length(Z),2)]
z2 <- fft(Z2, inverse=T)
plot(z)
points(Re(z2), col="red")
legend("topright", legend=c("original", "half-sampled"), fill=c("black", "red", "blue"))

# Second plot
x <- seq(1,0.101, -0.001)
X <- fft(x)/length(x)
X2 <- X[seq(1,length(X),2)]
x2 <- fft(X2, inverse=T)
plot(x)
points(Re(x2), col="red")     
legend("topright", legend=c("original", "half-sampled"), fill=c("black", "red", "blue"))

Answer 1

Regarding example 1:
first of all, either the fft or the ifft needs to be normalised by the number of sampling points as you have done (actually you can normalise each by a factor of the square root of the number of points, it is just a meter of definition). However, in your case the ifft is half the length of the fft you performed. Hence your signal is divided by 2. If you would perform the normalisation when calling the ifft (using half the original number of samples) you should get the correct height.
Now, regrading the second part of the signal, it is not truncated. What you see is some kind of aliasing. Since you downsampled without filtering, you get aliasing. This is the same concept as frequency aliasing (DFT and IDFT are basically the same). You should try to flip the input signal in the time domain and witness the results.

Next, regarding example 2:
Your problem is with the number of sampling points. Let me start by mentioning that since your signal is real, the frequency domain is symmetric.
In your first example you had even number of points. when you downsampled the frequency you still got a symmetric signal. In the second example you have odd number of samples. Therefore your decimated frequencies are not symmetric and the result is a complex signal. If you try it with one less (or more) sample it should give a similar result to example 1. On the other hand you can perform the frequency decimation while keeping the signal symmetric.