# Sampling with an alternating impulse train

The have the following question:

A signal $$m(t)$$ with bandwidth 500Hz is first multiplied by a signal $$g(t)$$ where $$\displaystyle g(t)=\sum_{k=-\infty}^{\infty}(-1)^k \delta(t-0.5*10^{-4}k)$$. The signal is then passed through an ideal low pass filter with bandwidth $$1$$kHz. The output of the low pass filter would be?

The solution is given in the below image: I am slightly confused with $$G(f)$$. As we know after sampling of a message signal $$x(t)$$ with frequency $$f_m$$ with the sampling rate being $$f_s$$, the output of the sampled signal $$x_s(t)$$ should have the following frequency components:

$$nf_s \pm f_m$$

Where $$n=0, 1, 2, ....$$

Then why according to the solution given there is nothing at $$n=0$$ i.e. $$0 \times f_s \pm f_m=\pm f_m$$, why there are no $$\pm f_m$$ components in the sampled signal or in other words why there in no component at $$f=0$$ for $$G(f)$$?

I was wondering if something is different due to the alternating impulse train.

We are not simply sampling the message in this case but also multiplying by a sampled tone that exists at $$f_s/2$$, (where $$f_s$$ is the sampling rate) and therefore we are frequency shifting the signal to this new location. This is intuitively clear from the two plots above. In the first case, representative of a sampling process, we are multiplying our waveform by an impulse train in time. Such an impulse train in time is a impulse train in frequency, with each impulse separated by 1/T where T is the time spacing between the impulses in time. Multiplication in time is convolution in frequency, so the process replicates the spectrum of the waveform at each of the impulses shown (as you showed you understood from the sampling process). Given the impulse that exists at frequency location 0, there will be a copy of the waveform spectrum here that is unchanged from the original, assuming no aliasing occurs from the other replicas at the other impulse locations in frequency (convolving with an impulse at freq = 0 does nothing).

In the lower plot we are doing something similar, but in addition we are multiplying the waveform by a alternating +/-1 sequence. Note that such a sequence in time has no DC value! The plot is very clear if you consider the time domain waveform to be a sampled cosine waveform, with a frequency that is exactly half the sampling period T. So in this case the waveform spectrum will be shifted away from the DC location; to each of the impulses as shown.

As the OP has found, see this link that further details the mathematics: Expression to represent alternating pulse train?

• Thanks a lot! Your explanation and maths from this answer helped me clear my confusion. If you feel like, you may add it as a reference :-) – paulplusx Nov 3 '18 at 14:44

In addition to Dan's excellent answer, which gives some intuition and a nice graphical representation of what is going on, I would like to give a concise mathematical derivation that might also be illuminating from a different viewpoint.

The key is to note that the given signal $$g(t)$$ can be represented as a sum of two standard impulse trains:

\begin{align}g(t)&=\sum_k(-1)^k\delta(t-kT/2)\\&=\sum_k\delta(t-kT)-\sum_k\delta(t-T/2-kT)\\&=g_1(t)-g_1(t-T/2)\tag{1}\end{align}

The Fourier transform of $$g_1(t)$$ is

$$G_1(f)=\frac{1}{T}\sum_k\delta(f-kf_s)\tag{2}$$

with $$f_s=1/T$$. From $$(1)$$ and $$(2)$$ we obtain the Fourier transform of $$g(t)$$ as

$$G(f)=\left(1-e^{-j\pi f/f_s}\right)\frac{1}{T}\sum_k\delta(f-kf_s)\tag{3}$$

The term in parentheses on the right-hand side of $$(3)$$ equals zero for $$f=2kf_s$$, and it equals $$2$$ for $$f=(2k+1)f_s$$. Consequently, $$G(f)$$ can be written as

$$G(f)=\frac{2}{T}\sum_k\delta(f-(2k+1)f_s)\tag{4}$$

Hence, multiplying a signal by $$g(t)$$ only generates images at odd multiples of $$f_s$$, and, consequently, there is no image centered at $$f=0$$.