1. CT-321 Digital Signal Processing
Yash Vasavada
Autumn 2016
DA-IICT
Lecture 3
DTFT
5th August 2016

2. Review and Preview Review of past lecture: Preview of this lecture:
Different types of signals
Frequency in continuous time (C-T) domain versus discrete time (D-T)
Several special D-T sequences: unit impulse, unit pulse, rectangular, exponential (real and complex)
Preview of this lecture:
Frequency domain representations
Rectangular pulse and time frequency duality
Discrete Time Fourier Transform or DTFT
First homework is posted on the web
Due next Friday

3. Several Special Sequences
Complex Exponential Sequence: x 𝑛 = 𝑎 𝑛 𝑒 𝑗 2𝜋𝑓𝑛+𝜃 = 𝑎 𝑛 cos 2𝜋𝑓𝑛+𝜃 +𝑗 sin 2𝜋𝑓𝑛+𝜃
Two representations of any complex sequence 𝑥 𝑛 :
(Cartesian) comprised of in-phase and quadrature phase components: 𝑥 𝑛 = 𝑥 𝐼 𝑛 +𝑗 𝑥 𝑄 (𝑛)
(Polar) Composed of magnitude and phase components: 𝑥 𝑛 = 𝑥 𝑛 𝑒 𝑗∠𝑥(𝑛)
Relationship between the two representations:
Magnitude is given as 𝑥 𝑛 = 𝑥 𝐼 2 𝑛 + 𝑥 𝑄 2 (𝑛)
Phase is given as ∠𝑥 𝑛 = tan −1 𝑥 𝑄 (𝑛)/ 𝑥 𝐼 (𝑛)
For example, when 𝑎=1, x 𝑛 = 𝑎 𝑛 𝑒 𝑗 2𝜋𝑓𝑛+𝜃 = 𝑒 𝑗 2𝜋𝑓𝑛+𝜃 has the following representations:
In-phase / quadrature phase: 𝑥 𝑛 = cos 2𝜋𝑓𝑛+𝜃 +𝑗 sin 2𝜋𝑓𝑛+𝜃
Magnitude / phase: 𝑥 𝑛 =1; ∠𝑥 𝑛 =2𝜋𝑓𝑛+𝜃

4. Several Special Sequences
Complex Exponential Sequence: x 𝑛 = 𝑎 𝑛 𝑒 𝑗 2𝜋𝑓𝑛+𝜃 = 𝑎 𝑛 cos 2𝜋𝑓𝑛+𝜃 +𝑗 sin 2𝜋𝑓𝑛+𝜃
In-phase sequence 𝑥 𝐼 𝑛 =Real{𝑥(𝑛)}
Quadrature phase sequence 𝑥 𝑄 (𝑛)=Imag{𝑥(𝑛)} 𝑛 𝑛
𝑎=0.8;𝑓=0.125

5. Several Special Sequences
Complex Exponential Sequence: x 𝑛 = 𝑎 𝑛 𝑒 𝑗 2𝜋𝑓𝑛+𝜃 = 𝑎 𝑛 cos 2𝜋𝑓𝑛+𝜃 +𝑗 sin 2𝜋𝑓𝑛+𝜃
Magnitude x n
Phase ∠𝑥(𝑛) (varies linearly over [−𝜋,+𝜋]) 𝑛 𝑛
𝑎=0.8;𝑓=0.125

6. Several Special Sequences: Complex Phasor
Phasor: x 𝑛 = 𝑒 𝑗 2𝜋𝑓𝑛+𝜃 = cos 2𝜋𝑓𝑛+𝜃 +𝑗 sin 2𝜋𝑓𝑛+𝜃 , 𝑎=1
Magnitude x n
Phase ∠𝑥(𝑛)
Time Domain Representation

7. Frequency Domain Representation
One more representation of sequence 𝑥(𝑛) = 𝑎 𝑛 𝑒 𝑗 2𝜋𝑓𝑛+𝜃 : frequency domain representation
Phasor: x 𝑛 = 𝑒 𝑗 2𝜋𝑓𝑛+𝜃 = cos 2𝜋𝑓𝑛+𝜃 +𝑗 sin 2𝜋𝑓𝑛+𝜃 , 𝑎=1
Magnitude x n
Phase ∠𝑥(𝑛)
Complex phasor at 𝑓=0.125
𝑒 𝑗𝜃 𝛿(𝑓−0.125)
Domain Transformation 𝑓
0.125
−0.5
+0.5
Time Domain Representation
Frequency Domain Representation

8. Frequency Domain Representation
Message from the prior slide:
Although a signal may appear to exhibit a lot of “data” in one domain, it might actually be a very simple signal when looked at in the transformed domain
All you need to specify a phasor are its
Location in frequency domain
Amplitude 𝐴 and
Starting phase 𝜃
A digital signal processor can generate as many number of time domain samples of this phasor as desired given just this information
Such a DSP is called NCO – Numerically Controlled Oscillator
This is a recurrent theme in the application of DSP to achieve data compression.
Transform the domain to convert “big data” into small, easily manageable, compressed data, and vice versa

9. Frequency Domain Representation
Frequency domain representations of (continuous) time domain signals
Reference:
Time domain and frequency domain columns can be flipped (with small changes) due to the duality relation
Therefore, oftentimes, a complicated-looking signal in one domain may be a simple signal in another domain

10. Frequency Domain Representations
Rectangular Pulse and its Frequency Domain Representation*
Time Domain: Π 𝑡 𝑇 =𝐴, 𝑡 ≤ 𝑇 2 ; 0 otherwise Frequency Domain: A𝑇 sinc 𝑓𝑇
𝒕 →
𝒇 →
*Reference:

11. Frequency Domain Representations
Time/frequency duality: localization in one domain results in a spread in another domain
Rectangular Pulses and its Frequency Domain Representations* shown for 𝑇=1 and 𝑇=10
Consider what would happen if 𝑇 is made arbitrarily small, or arbitrarily large
Time Domain Frequency Domain
Time Domain Frequency Domain
𝑇=1
𝑇=10

12. Frequency Domain Representations
Rectangular pulse is a key signal of frequent application in theoretical DSP studies
To see why, consider a hypothetical signal that we have received, that has an undesired component mixed in with the desired signal. We want to get rid of the unwanted component.
One solution is to multiply this received signal by a rectangular pulse of unity amplitude and of sufficient width centered at the desired signal’s location.
This will zero out the unwanted component and leave the desired part unchanged.
Notice that this mixture of desired and unwanted signals can occur either in time domain or in frequency domain
In time domain, it’s not hard to envision implementation of the multiplication operation between the received signal and the rectangular pulse. However, how to do so in the frequency domain? 1
Unwanted
Desired
Time or Frequency

13. Rectangular Pulse in Discrete Time Domain
We have so far looked at the frequency domain representation of the rectangular pulse in continuous time domain. What about discrete time domain?
A useful mathematical series for DSP engineers: a sum of (2𝑁+1) samples of a phasor rotating at frequency 𝐹 from sample index −𝑁 to 𝑁
How to derive a closed-form expression of 𝑆 𝐹 ?  Begin by writing 𝑆(𝐹) as follows:
Closed form expression of 𝑆 𝐹 :
Surprisingly quite similar to the Fourier Transform of the rectangular pulse in continuous time domain!
Why?
Now solve using this expression:

14. Discrete Time Fourier Transform
Discrete Time Fourier Transform (DTFT) allows us to obtain frequency domain representation of discrete time signals
Discrete time analogue of the conventional (continuous time) Fourier Transform
Defined as follows:
Several items to note:
DTFT is evaluated for a given value of frequency 𝑓.
What is the range of this frequency 𝑓? Is 𝑓 continuous (analog) or discrete?
What is the DTFT of the rectangular pulse that we have looked at earlier?

15. DTFT of the Rectangular Pulse
(Frequency Domain Representation of Rectangular Pulses)
Time Domain Rectangular Pulses
Frequency domain representation becomes localized (is narrower) as the time domain signal spreads out (becomes wider)
Same as the observation we have made earlier in context of C-T signals
Can DTFT be reversed?
i.e., can the discrete time samples be recovered from continuous- frequency DTFT?
If so, what is likely to be this “inverse” DTFT of rectangular function in the frequency domain?
𝑁=2
Peak value is given as 2𝑁+1. Why?
Locations of the first nulls are always at 𝑓= ±1 2𝑁+1
𝑁=4
Values at the extreme points of 𝑓=± 1 2 are always the same and they are either +1 or −1 𝑓