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

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

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

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

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

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

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

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

Frequency Domain Representation Frequency domain representations of (continuous) time domain signals Reference: http://uspas.fnal.gov/materials/11ODU/FourierTransformPairs.pdf 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

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: http://www.thefouriertransform.com/pairs/box.php

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

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

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:

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?

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