Ideal Sampling

In order to work with continuous signals using a computer, we need to sample them. This means recording the value at particular points of time. During uniform sampling, we take samples at an even sampling period
is our continuous signal). This is done by passing the signal through an Analog-To-Digital converter. From there we can do discrete time processing and reconstruct our signal by passing it through a Digital-to-Analog converter with reconstruction period
Figure 6: Uniform Sampling System
We mathematically model sampling as multiplication by an impulse train. Notice that if we were to take a signal
and multiply it by an impulse train, then we would get a series of impulses equal to
at the sampling points and
everywhere else. We can call this signal
p(t)=k=δ(tkT)p(t) = \sum_{k=-\infty}^{\infty}{\delta(t-kT)}
xp(t)=x(t)p(t)=k=x(t)δ(tkT)x_p(t) = x(t)p(t) = \sum_{k=-\infty}^{\infty}{x(t)\delta(t-kT)}
In the Fourier Domain,
Xp(jΩ)=12πX(jΩ)P(jΩ)P(jΩ)=2πTk=δ(ΩkΩs)Xp(jΩ)=12πX(jθ)P(j(Ωθ))dθ=1Tk=X(j(ΩkΩs)\begin{aligned} X_p(j\Omega) &= \frac{1}{2\pi}X(j\Omega)*P(j\Omega)\\ P(j\Omega) &= \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}{\delta(\Omega-k\Omega_s)}\\ \therefore X_p(j\Omega) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}{X(j\theta)P(j(\Omega-\theta))d\theta} = \frac{1}{T}\sum_{k=-\infty}^{\infty}{X(j(\Omega-k\Omega_s)}\end{aligned}
What this tells us is that the Fourier Transform of our sampled signal is a series of copies of
, each centered at
Ωs=2πT\Omega_s = \frac{2\pi}{T}
. This is a good model because we can equivalently write the CTFT of the impulse train sampled signal as
Xp(jΩ)=k=x(t)δ(tkT)k=x(kT)ejkTΩ.\begin{aligned} X_p(j\Omega) &= \int_{-\infty}^{\infty}\sum_{k=-\infty}^{\infty}{x(t)\delta(t-kT)} \sum_{k=-\infty}^{\infty}x(kT)e^{-jkT\Omega}.\end{aligned}
Notice that this is just the DTFT of
if we set
ω=ΩT\omega = \Omega T
X(ejω)=n=x(nT)ejωn=Xp(jΩ)Ω=ωT=1Tk=X(ωTk2πTs)X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x(nT)e^{-j\omega n}=X_p(j\Omega)|_{\Omega=\frac{\omega}{T}}=\frac{1}{T}\sum_{k=-\infty}^{\infty}{X\left(\frac{\omega}{T}-k\frac{2\pi}{T_s}\right)}
This means that the DTFT of our signal is just a bunch of shifted copies, and the frequency axis is scaled so
Ωs2π\Omega_s \rightarrow 2\pi

Nyquist Theorem

To analyze this further, we will stay in continuous time. Lets say that our original signal has the following Fourier Transform. Notice the signal is band-limited by
Figure 7: Example of the spectrum of a bandlimited signal
There are two major cases: if
Ωs>2Ωm\Omega_s > 2\Omega_m
Ωs<2Ωm\Omega_s < 2\Omega_m
Case One:
Ωs>2Ωm\Omega_s > 2\Omega_m
Figure 8: When $\Omega_s > 2\Omega_m$
As shown in Figure 8, the shifted copies of the original
(shown in blue) do not overlap with each other or with the original copy. If we wanted to recover the original signal, we could simply apply a low pass filter to isolate the unshifted copy of
and then take the inverse Fourier Transform.
Case Two:
Ωs<2Ωm\Omega_s < 2\Omega_m
Figure 9: When $\Omega_s < 2\Omega_m$
Notice how in Figure 9, the shifted copies overlap with the original
. This means in our sampled signal, the higher frequency information is bleeding in with the lower frequency information. This phenomenon is known as aliasing. When aliasing occurs, we cannot simply apply a low pass filter to isolate the unshifted copy of
Ωs=2ΩM\Omega_s = 2\Omega_M
, then our ability to reconstruct the original signal depends on the shape of its Fourier Transform. As long as
are equal to
), then we can apply an LPF because we can isolate the original
and take its inverse Fourier Transform. Remember that an ideal low pass filter is a square wave in the frequency domain and a
in the time domain. Thus if we allow
Xr(jΩ)=Xp(jΩ){TΩ<Ωs20 else }X_r(j\Omega) = X_p(j\Omega)\cdot \left\{ \begin{array}{cc} T & |\Omega| < \frac{\Omega_s}{2}\\ 0 & \text{ else } \end{array} \right\}
then our reconstructed signal will be
xr(t)=xp(t)sinc(tT)=n=X(nT)sinc(tnTT).x_r(t) = x_p(t)*\text{sinc}\left(\frac{t}{T}\right) = \sum_{n=-\infty}^{\infty}{X(nT)\text{sinc}\left(\frac{t-nT}{T}\right)}.
This is why we call reconstructing a signal from its samples "sinc interpolation." This leads us to formulate the Nyquist Theorem.

Theorem 4 (Nyquist Theorem)

Suppose a continuous signal
is bandlimited and we sample it at a rate of
Ωs>2Ωm\Omega_s > 2\Omega_m
, then the signal
reconstructed by sinc interpolation is exactly

Discrete Time Processing of a Continuous Time Signal

As long as the DT system we apply is LTI, the overall CT system will be linear too, but it will not necessarily be time invariant because sampling inherently depends on the signal’s timing. If we want to find the overall CT transfer function (
ω=ΩT\omega = \Omega T
) of a system like that depicted in Figure 6.
Yd(ejω)=Hd(ejω)Xd(ejω)=Hd(ejω)Xp(ωT)Yp(jΩ)=Yd(ejΩT)=Hd(ejΩT)Xp(jΩ)Y(jΩ)={TΩ<Ωs20ΩΩs2}Yp(jΩ)={THd(ejΩT)Xp(jΩ)Ω<Ωs20ΩΩs2}\begin{aligned} Y_d(e^{j\omega}) &= H_d(e^{j\omega})X_d(e^{j\omega}) = H_d(e^{j\omega})X_p\left(\frac{\omega}{T}\right)\\ Y_p(j\Omega) &= Y_d(e^{j\Omega T}) = H_d(e^{j\Omega T})X_p(j\Omega)\\ Y(j\Omega) &= \left\{ \begin{array}{cc} T & |\Omega| < \frac{\Omega_s}{2}\\ 0 & |\Omega| \ge \frac{\Omega_s}{2} \end{array} \right\} \cdot Y_p(j\Omega) = \left\{ \begin{array}{cc} TH_d(e^{j\Omega T})X_p(j\Omega) & |\Omega| < \frac{\Omega_s}{2}\\ 0 & |\Omega| \ge \frac{\Omega_s}{2} \end{array} \right\}\end{aligned}
Assuming that the Nyquist criteria is met holds,
Xp(jΩ)=1TX(jΩ)Y(jΩ)={Hd(ejΩT)X(jω)Ω<Ωs20ΩΩs2}Hsystem={Hd(ejωT)Ω<Ωs20ΩΩs2}\begin{aligned} X_p(j\Omega) &= \frac{1}{T}X(j\Omega)\\ \therefore Y(j\Omega) &= \left\{ \begin{array}{cc} H_d(e^{j\Omega T})X(j\omega) & |\Omega| < \frac{\Omega_s}{2}\\ 0 & |\Omega| \ge \frac{\Omega_s}{2} \end{array} \right\}\\ \therefore H_{system} &= \left\{\begin{array}{cc} H_d(e^{j\omega T}) & |\Omega| < \frac{\Omega_s}{2}\\ 0 & |\Omega| \ge \frac{\Omega_s}{2} \end{array} \right\}\end{aligned}
This shows us that as long as the Nyquist theorem holds, we can process continuous signals with a disrete time LTI system and still have the result be LTI.

Continuous Time Processing of Discrete Time Signals

While not useful in practice, it can be useful to model a discrete time transfer function in terms of Continuous Time processing (e.g a half sample delay).
Figure 10: Continuous Time processing of a Discrete Time signal
Similar to the analysis of DT processing of a CT signal, we can write the discrete transfer function in terms of the continuous function. Our continuous signal will be bandlimited after reconstruction.
X(jΩ)={TXd(ejω)ω=ΩTΩΩs20X(j\Omega) = \begin{cases} T X_d(e^{j\omega})|_{\omega=\Omega T} & |\Omega| \le \frac{\Omega_s}{2}\\ 0 \end{cases}
This means our reconstructed signal
is also bandlimited, so we can say that


When we downsample a signal by a factor of
, we create a new signal
by taking every
sample. What this means conceptually is that we are reconstructing the continuous signal and then sampling it at a slower rate
was the original sampling rate. If
is the original continuous time signal and
is the sampled signal, then the downsampled signal
will be
y[n]=x[nM]=xc(nMT)    Y(ejω)=1MTk=Xc(ωNTk2πNT).y[n]=x[nM]=x_c(nMT)\implies Y(e^{j\omega}) =\frac{1}{MT}\sum_{k=-\infty}^{\infty}X_c\left(\frac{\omega}{NT}-k\frac{2\pi}{NT}\right).
If we re-index and let
m[0,N1],pZm\in [0, N-1],p\in \mathbb{Z}
Y(ejω)=1Mm=0M1Xd(ejω2πmM).Y(e^{j\omega})=\frac{1}{M}\sum_{m=0}^{M-1}X_d(e^{j\frac{\omega-2\pi m}{M}}).
What this means is to obtain the new DTFT, we need to scale the frequency axis so
πMπ\frac{\pi}{M}\rightarrow \pi
. To prevent aliasing when this happens, we include an LPF before the downsample step.
Figure 11: Downsampling


When we upsample a signal by a factor of L, we are interpolating between samples. Conceptually, this means we are reconstructing the original continuous time signal and resampling it at a faster rate than before. First we place zeros in between samples, effectively expanding our signal.
xe[n]={x[nL]n=0,±L,±2L,...0x_e[n] = \begin{cases} x\left[\frac{n}{L}\right] & n=0, \pm L, \pm 2L,...\\ 0 \end{cases}
Xe(ejω)=xe[n]ejωn=m=x[m]ejωmL=X(ejωL)X_e(e^{j\omega})=\sum_{-\infty}^{\infty}x_e[n]e^{-j\omega n}=\sum_{m=-\infty}^{\infty}x[m]e^{-j\omega mL} = X\left(e^{j\omega L}\right)
Then we interpolate by convolving with a
y[n]=xe[n]sinc(nL)=n=x[k]sinc(nkLL)y[n] = x_e[n]*\text{sinc}\left(\frac{n}{L}\right) = \sum_{n=-\infty}^{\infty}{x[k]\text{sinc}\left(\frac{n-kL}{L}\right)}
In the frequency domain, this looks like compressing the frequency axis so
ππL\pi \rightarrow \frac{\pi}{L}
and then taking a low pass filter.
Figure 12: Upsampling operation
The gain of L is used to scale the spectrum so it is identical to if we had sampled the continuous signal at a rate of

Multi-Rate Signal Processing

In order to resample a signal to a rate
where T is the original sampling rate, we can do this by upsampling then downsampling our signal.
Notice that we only need one LPF to take care of both anti-aliasing and interpolation.

Exchanging Filter Order During Resampling

Notice that resampling with a very small change wastes a lot of computation. For example, resampling with
would upsample by 100 and then throw away most of those samples when we downsample. Thus it would be useful to exchange the order of operations when resampling to save computation.
Figure 13: Interchanging an upsampling operation
During upsampling, we convolve our filter with a bunch of zeros caused by the expansion. Convolution with 0’s is a unnecessary, so instead we could convolve with a compressed version of the filter. Notice the results will be the same as long as
is a rational function,
Figure 14: Interchanging a downsampling operation
During downsampling, we do a convolution and then throw away most of our results. It would be much more efficient to instead compute only the quantities we need. This is accomplished by downsmapling first and then convolving. Just like before, the results are only going to be the same if
is a rational function.

Polyphase Decomposition

The problem with interchanging filters is that it is not always possible. Most filters are not compressible. However, we can get around this issue and still get the efficiency gains of interchanging filter orders by taking a polyphase decomposition of our filters. First notice that
can be written as a sum of compressible filters.
h[n]=k=0M1hk[nk]h[n] = \sum_{k=0}^{M-1}h_k[n-k]
Figure 15: Example of decomposing a filter: M=2
This means if we let
ek[n]=hk[nM]e_k[n] = h_k[nM]
, we can utilize the linearity of convolution to build a bank of filters.
Figure 16: A filter bank
Now each of our filters is compressible, so we can switch the order of downsampling and filtering while maintaining the same output.
Figure 17: Filter bank but with the downsampling done first
Now for any filter, we can compute only what we need, so the result is correct and efficently obtained.

Practical Sampling (ADC)

Unfortunately, ideal analog to digital conversion is not possible for a variety of reasons. The first is that not all signals are bandlimited (or there may be noise outside of the bandwidth). Moreover, computers only have finite precision, so we cannot represent the full range of values that a continuous signal might take on with a finite number of bits per sample. The solution to the first issue is to include a “anti-aliasing” filter before the sampler. The solution to the second issue is to quantize.
Figure 18: Sampling with quantization
However, sharp analog filters are difficult to implement in practice. To deal with this, we could make the anti-aliasing filter wider, but this would add noise and interference. If we keep the cutoff frequency the same, then we could alter part of the signal because our filter is not ideal. A better solution is to do the processing in Discrete Time because we have more control. We also sample higher than the Nyquist Rate and then downsample it to the required rate.
Figure 19: A practical sampling system with quantization and anti-aliasing


If we have a dynamic range of
is the length of the range of values we can represent), then our step between quantized values is
, assuming we are representing our data as 2’s complement numbers with
bits. We model the error caused by quantization as additive noise. Our quantized signal
is decribed by
x^[n]=x[n]+e[n]Δ2e[n]Δ2\hat{x}[n] = x[n] + e[n] \qquad \frac{-\Delta}{2}\le e[n] \le \frac{\Delta}{2}
We do this under the following assumptions:
  1. 1.
    is produced by a stationary random process
  2. 2.
    is not correlated with
  3. 3.
    is white noise (
    is not correlated with
  4. 4.
    e[n]U[Δ2,Δ2]e[n]\sim U\left[\frac{-\Delta}{2},\frac{\Delta}{2}\right]
For rapidly changing signals with small
, this assumptions hold, and they are useful in modeling quantization error. Since
Δ=2BXm\Delta = 2^{-B}X_m
This means our Signal to Noise Ratio for quantization is
What this tells us is that every new bit we add gives us 6dB in improvement. It also tells us that we need to adapt the range of quantization to the RMS amplitude of the signal. This means there is a tradeoff between clipping and quantization noise. When we oversampling our signal, we can further limit the effects of quantization noise because this noise will be spread out over more frequencies and the LPF will eliminate noise outside the signal bandwidth. This makes
the new noise variance (if we oversample by
). Thus we can modify the
SNRQ=6.02B+10.820log(Xmσs)+10logM.SNR_Q=6.02B+10.8-20\log\left(\frac{X_m}{\sigma_s}\right) + 10\log M.
This shows that doubling
yields a 3dB improvement (equivalent to 0.5 more bits).

Practical Reconstruction (DAC)

In the ideal case, we reconstruct signals by converting them to impulses and then convolving with a sinc. However, impulses are require lots of power to generate, and sincs are infinitely long, so it is impractical to design an analog system to do this. Instead, we use an interpolation like Zero-Order-Hold to pulses and then filter with a reconstruction filter.
Figure 20: Practical Reconstruction
Xr(jΩ)=Hr(jΩ)TejΩT2sinc(ΩΩs)Zero Order Hold1Tk=X(j(ΩkΩs))Sampled SignalX_r(j\Omega) = H_r(j\Omega)\overbrace{Te^{-j\Omega\frac{T}{2}}\text{sinc}\left(\frac{\Omega}{\Omega_s}\right)}^{\text{Zero Order Hold}}\overbrace{\frac{1}{T}\sum_{k=-\infty}^{\infty}X(j(\Omega-k\Omega_s))}^{\text{Sampled Signal}}
We design
such that
is approximately an LPF.