Section 8: Carrier Modulated Transmission Convert binary data into - PowerPoint PPT Presentation

8.3.5. Coherent versus non-coherent detection Let the signal be: b(t)cos(2 πƒ c t). Noise is: N(t)cos(2 πƒ c t + θ (t)) where N(t) is random envelope & θ (t) is random phase. This equals: θ π + θ π N ( t ) cos ( t ) cos( 2 f t ) N ( t ) sin ( t ) sin( 2 f t ) c c Half noise power in phase with cos(2 πƒ c t ) & half with sin(2 πƒ c t ). Non-coherent detection measures envelope of signal plus noise & is affected by full power of noise. Coherent detection multiplies by cos(2 πƒ c t ) low-pass filters & thus eliminates half the noise power 3dB reduction in effective noise power as seen by detector. ∴ coherent detection tolerates 3dB more noise than non-coherent to achieve same BER. 24 Apr'06 CS3282 Sectn 8 37

8.4 Complx baseband & vector-modulator/demodulatr 8.4.1 Vector modulator: sin(2 π f C t ) b I (t) ..11010. . Map b R (t)cos(2 π f C t ) + b I (t)sin(2 π f C t ) b R (t) ..10010.. Map cos( 2 π f C t ) 24 Apr'06 CS3282 Sectn 8 38

Complex notation for vector-modulator • b R (t) is ‘in-phase’ component & b I (t) is ‘quadrature’ component. • Complex base-band signal is b R (t) + j b I (t) where j = √ (-1). • Output is real part of: [ b R (t) + j b I (t)] . exp(-2 π jf C t ) since [ b R (t) + j b I (t)] . [ cos(2 π f C t ) − j sin(2 π f C t ) ] = [ b R (t) cos(2 π f C t ) + b I (t)sin(2 π f C t ) ] + j (..) b(t) 10110 Complex Map Mult signal. Take Complx real part. 11011 base-band exp(-2 π j f C t) 24 Apr'06 CS3282 Sectn 8 39

8.4.2. Vector-demodulator • Receives b R (t)cos(2 π f C t ) + b I (t)sin(2 π f C t ) • Recovers b R (t) & b I (t) separately. • b R (t) & b I (t) may be considered independent channels. • If each transmits at 1 b/s/Hz, we get 2 b/s per Hz. • “Two channels for price of one”. • Constellation diagrams becomes more interesting: 24 Apr'06 CS3282 Sectn 8 40

Vector demodulator (cont) Sin(2 π f C t) b I (t) ..11010.. Low Threshold Mult pass Detector Received signal r(t) ..10010.. Threshold Low Mult Detector pass Derive local b R (t) carrier Cos(2 π f C t) (cos & sin) 24 Apr'06 CS3282 Sectn 8 41

Show why this works for cosine modulation Let r(t) = b R (t) cos(2 π f C t ) + b I (t) sin(2 π f C t ) ) Then r(t) cos(2 π f C t ) = b R (t)cos 2 (2 π f C t ) + b I (t) sin(2 π f C t ) )cos(2 π f C t ) = 0.5 b R (t)[1 + cos(4 π f C t )] + 0.5 b I (t) sin(4 π f C t ) ) = 0.5b R (t) + 0.5b R (t) cos(4 π f C t ) + 0.5 b I (t) sin(4 π f C t ) ) Removed by lowpass filter Hence cosine demodulator recovers b R (t) & is blind to b I (t) 24 Apr'06 CS3282 Sectn 8 42

Similarly for sine modulation r(t)sin(2 π f C t ) = b R (t) cos(2 π f C t )sin(2 π f C t ) + b I (t) sin 2 (2 π f C t ) ) = 0.5 b R (t) sin(4 π f C t ) + 0.5 b I (t) [1 - cos(4 π f C t ) ] = 0.5 b R (t) sin(4 π f C t ) + 0.5 b I (t) - 0.5b I (t)cos(4 π f C t ) Removed by lowpass filter Sine demodulator recovers b I (t) & is blind to b R (t) 24 Apr'06 CS3282 Sectn 8 43

Trig formulae This works because cos 2 ( θ ) & sin 2 ( θ ) have a constant (or DC) component 0.5 whereas sin( θ )cos( θ ) does not. Relevant formulae are: • cos 2 ( θ ) = 0.5 + 0.5 cos(2 θ ) • sin 2 ( θ ) = 0.5 - 0.5 cos(2 θ ) • sin( θ ) cos ( θ ) = 0.5sin(2 θ ) 24 Apr'06 CS3282 Sectn 8 44

8.4.3. Constellation diags for ASK with complx baseband Quadrature to In quadrature carrier A 3A In phase A with carrier A In 0 phas 0 A 2A 3A Binary ASK 4-ary ASK for b R (t) & b I (t) for b R (t) & b I (t) 24 Apr'06 CS3282 Sectn 8 45

Symbol allocation tables for binary & 4-ary ASK Bits b R b I 0 0 0 0 0 0 0 0 0 1 0 A Bits b R b I 0 0 1 0 0 2A 0 0 0 0 0 0 1 1 0 3A 0 1 0 A 0 1 0 0 A 0 1 0 A 0 0 1 0 1 A A 1 1 A A 0 1 1 0 A 2A 0 1 1 1 A 3A 1 0 0 0 2A 0 1 0 0 1 2A A ..... 1 1 1 1 3A 3A 24 Apr'06 CS3282 Sectn 8 46

8.5 Frequency Shift Keying (FSK) • Can be straightforward form of digital modulation. • Simple to generate and detect, • Constant amplitude, ∴ insensitive to fluctuations of channel attenuation. • Based on frequency modulation (fm) • Uses set of distinct frequencies to represent symbols. • Transmit constant amplitude sine-wave whose frequency varies between the frequencies assigned to each symbol. •For binary signalling there are 2 frequencies, ƒ 0 & ƒ 1 say. •Consider 3 generation methods. 24 Apr'06 CS3282 Sectn 8 47

8.5.1 Methods for generating FSK 1. “ Voltage controlled oscillator(VCO)” method. FM Modulator 1 (VCO) 0 1 0 0 Better to have smoothly changing pulse for gradual transition. This is “continuous phase form of FSK i.e. CPFSK. 2. “Switched oscillator ” method of generating FSK . Clearly this may not produce a FSK continuous phase 1 output. 0 24 Apr'06 CS3282 Sectn 8 48

3. “ Vector-modulator ” method: For binary FSK with ƒ c + ƒ 1 & ƒ c - ƒ 1 , apply cos (2 πƒ 1 t) to ‘Q’ and ± sin(2 πƒ 1 t) to ‘I’ . Sign determines the symbol. Sin(2 πƒ c t) “Q” input cos (2 πƒ 1 t) “I” input ± sin(2 πƒ 1 t) Cos(2 πƒ c t) 24 Apr'06 CS3282 Sectn 8 49

Exercise 8.1: Check that this works. Solution: When I=+sin(2 πƒ 1 t), output is: sin(2 πƒ 1 t)cos(2 πƒ c t)+cos(2 πƒ 1 t)sin(2 πƒ c t) =sin(2 π ( ƒ c + ƒ 1 )t) When I=-sin(2 π f 1 t) the output is: -sin(2 πƒ 1 t)cos(2 πƒ c t)+cos(2 πƒ 1 t)sin(2 πƒ c t) =sin(2 π ( ƒ c – ƒ 1 )t) 24 Apr'06 CS3282 Sectn 8 50

8.5.2. Non-coherent detection of FSK at receiver (low bit-rates) Consider 3 methods 1. Set of band-pass filters with envelope-detectors; BPF ( f 0) Decide BPF ( f 1) 24 Apr'06 CS3282 Sectn 8 51

2. Discriminator followed by envelope-detector. Turns FSK into ASK for easier detection V t t t t f 1 f 0 Low-pass filter (smoother) Discriminator Gain Resistor f f 1 f 0 24 Apr'06 CS3282 Sectn 8 52

3. Phase Locked Loop detector for FSK. PLL is 'black box' with one input & 2 useful outputs: V t t VCO input (Voltage ∝ input PLL frequency) Frequency VCO modulated input t output 24 Apr'06 CS3282 Sectn 8 53

8.5.3. Phase-locked loop (PLL) PLL has VCO with frequency adapted to match that of FSK signal. VCO controlled by voltage generated by measuring phase difference between VCO output & incoming FSK signal. Voltage ∝ input frequency & can be used for detecting data bits V t t VCO input Low-pass voltage filter V VCO VCO output voltage 24 Apr'06 CS3282 Sectn 8 54

8.5.4 Non-coherent FSK detector for higher data rates: “Zero crossing counter” type of detector Limiting Amplifier FSK Data and Counter Decide Clock Reset 24 Apr'06 CS3282 Sectn 8 55

8.5.5 Coherent FSK detection : Similar to coherent ASK detection. Must have local carrier sine-waves at receiver. Must match exactly in frequency & phase the FSK symbols being received. For binary transmission there would be two locally generated sine- waves of frequency ƒ 0 and ƒ 1 respectively. The incoming signal is multiplied by both sine waves and the two signals which result are low-pass filtered. A comparator then has to decide which frequency ƒ 0 or ƒ 1 produced the larger output, and that determines the symbol. 24 Apr'06 CS3282 Sectn 8 56

8.5.6 Spectrum of FSK: At 1/T symbols/s, base-band signal has spectrum which is non-zero for –1/T< ƒ <1/T if 100% RC spectral shaping is applied Non-zero for –1/(2T)< ƒ <1/(2T) with 0% RC spectral shaping. When base-band signal is modulated to form FSK with signalling frequencies ƒ 1 & ƒ 0 , ‘one’s form a DSB spectrum centred on ƒ 1 ‘zero’s form a DSB spectrum centred on ƒ 0 . Resulting spectrum is sum of these two spectra. PSD PSD ƒ ƒ ƒ 0 -1/T ƒ 0 ƒ 0 +1/T ƒ 1 -1/T ƒ 1 ƒ 1 +1/T 24 Apr'06 CS3282 Sectn 8 57

PSD PSD = + ƒ ƒ ƒ 0 -1/T ƒ 0 ƒ 0 +1/T ƒ 1 -1/T ƒ 1 ƒ 1 +1/T PSD ƒ ƒ 1 +1/T ƒ 0 -1/T ƒ 0 ƒ 1 24 Apr'06 CS3282 Sectn 8 58

Sunde’s FSK method Place ƒ 0 at ƒ 1 ± 1/T & ƒ 1 at ƒ o ± 1/T. 24 Apr'06 CS3282 Sectn 8 59

8.5.7. Minimum shift keying (MSK) •Form of FSK where difference between ƒ 0 & ƒ 1 is 1/(2T) Hz. •Makes MSK very efficient in its spectral utilisation. •Price is increased complexity in generation & detection process. •Non-coherent detection is difficult for MSK. •The detection is recommended to be coherent (Sklar p152). Pulse-shaping filter: e.g. 100r % RRC, controls FSK spectrum. •Placed just before the FSK modulator. •Controls how frequency changes from ƒ 0 to ƒ 1 and vice-versa. •In GSM phone systems the shaping is root-Gaussian filter. •This form of binary FSK is known as “ Gaussian MSK ”. 24 Apr'06 CS3282 Sectn 8 60

GMSK transmitter GMSK Map to FIR ..10110 impulse Gaussian .. VCO s shaping filter 24 Apr'06 CS3282 Sectn 8 61

Gaussian minimum shift keying (GMSK) • Spectrally efficient form of binary FSK with ‘Gaussian’ pulse shaping. ≈ 2 bits/s /Hz • • Spectrum similar to ASK • Used for GSM 24 Apr'06 CS3282 Sectn 8 62

8.5.8. Advantages & disadvantages of FSK Advantages: 1. Constant envelope hence not too sensitive to varying attenuation on the channel. 2. Detection based on frequency changes, so not very sensitive to frequency shifts of channel, (Doppler shifts etc). 3. Simple implementations possible for low bit-rates. Disadvantages of FSK: 1. Less bandwidth efficient than ASK or PSK (except MSK) 2. Bit-error rate performance in AWGN worse than PSK. 24 Apr'06 CS3282 Sectn 8 63

8.6. Phase shift keying (PSK) • Send sinusoidal carrier with phase changes determined by bits • Consider binary PSK with 1 bit/cycle, 0 0 & 180 0 phase shifts & rectangular pulse shaping b(t) t ..1010010.. Map ± cos(2 πƒ c t) cos(2 πƒ c t) 24 Apr'06 CS3282 Sectn 8 64

A binary PSK waveform V t 1 1 0 0 1 1 0 Assuming 1 bit per cycle. 24 Apr'06 CS3282 Sectn 8 65

8.6.2 Coherent Detector for binary PSK ± cos 2 (2 πƒ c t) Data = ± 0.5(1+cos4 πƒ c t) +1/2:”1” ± cos(2 πƒ C t) -1/2:”0” Lowpass Threshold filter Detector ± 1/2 Generate cos(2 πƒ C t) local carrier 24 Apr'06 CS3282 Sectn 8 66

Details of coherent PSK demodulator/detector •Low-pass filter eliminates ± cos(4 πƒ C t). •Matched filter will achieve this because of orthogonality of ± cos(4 πƒ c t) to sin(2 πƒ c t). •Local carrier must be generated from received signal. (Square incoming signal & divide frequency of result by 2). •Spectrum of PSK similar to that of ASK. •PSK multiplies carrier by bipolar base-band: ASK by unipolar. •Shifts up base-band spectrum producing DSB spectrum centred on carrier frequency. 24 Apr'06 CS3282 Sectn 8 67

8.6.3 Differential PSK Phase shift of carrier with respect to previous symbol indicates current bit. Illustrate for 1 bit/cycle with 0 0 shift for 1 & 180 0 for 0: V t 1 0 1 0 0 0 (0 o ) (180 0 ) 24 Apr'06 CS3282 Sectn 8 68

90 0 & 270 0 phase shifts often preferred with binary DPSK : V t 1 bit/cycle 1 1 0 1 1 0 Discontinuities tell receiver when next symbol starts. Makes bit-synchronisation easier when symbol rate not fully synchronised with carrier (not exact no. of cycles/bit).. 24 Apr'06 CS3282 Sectn 8 69

8.6.4 Differential detection of binary DPSK •Consider case where phase shifts are 0 0 & 180 0 & there is an integer number (e.g. 1) of cycles per bit. •Instead of generating local carrier, take previous symbol delayed as required carrier segment. •Small penalty compared with a fully coherent technique. ± cos 2 (2 πƒ c t) ± 0.5 = ± 0.5(1+cos4 πƒ c t) ± cos(2 πƒ C t) Lowpass Threshold filter detector Delay by T (Delay for 1 bit) 24 Apr'06 CS3282 Sectn 8 70

Lowpass filter output is +0.5 if carrier has been subject to 0 0 phase shift (logic 1 say) and –1/2 for 180 0 (logic ‘0’). Channel noise affects both data & delayed data used as carrier. Was used for modem data over telephone lines, 1200 b/s being possible over worst case lines. Increased to 2400bits/s using quaternary PSK (QPSK). 24 Apr'06 CS3282 Sectn 8 71

8.6.5 Detector for binary DSPK with 90 O & 270 O phase shifts rather than 0 and 180 O . LPF Detect Delay by T 90 0 phase (Delay for 1 shift bit) 24 Apr'06 CS3282 Sectn 8 72

8.6.6 Quaternary PSK (QPSK) • Consider a vector modulator where b R (t) & b I (t) are bipolar • Then b R (t)cos(2 π f C t) & b I (t) sin(2 π f C t) are both binary PSK. • ‘2-channel’ modulation process is QPSK or 4-PSK. Sin(2 π f C t) 10110 Map b I (t) Mult ADD Mult Map b R (t) 11011 Cos(2 π f C t) 24 Apr'06 CS3282 Sectn 8 73

QPSK de-modulator Sin(2 π f C t) b I (t) 10110 Low Detect Mult pass 11011 Low Mult Detect pass Detect b R (t) Cos(2 π f C t) carrier 24 Apr'06 CS3282 Sectn 8 74

Two ways of looking at QPSK • One way is ‘vector modulation’ approach where cos(2 π f C t) & sin(2 π f C t) are binary PSK modulated independently. • At receiver, coherent PSK detector for cos(2 π f C t) channel is blind to transmission on sin(2 π f C t) & vice-versa. • Refer to b R (t) + j b I (t) as 'complex base-band' signal b(t). • Transmitted QPSK signal is Re{ [b R (t) + j b I (t)] exp(-j2 π f C t) }. 10110 b(t) Map Transmit Mult real part Complx 11011 base-band exp(-2 π j f C t) 24 Apr'06 CS3282 Sectn 8 75

Another way to look at QPSK • QPSK sends 2 bits at once , using bipolar b R (t) & b I (t) • Let b R (t) & b I (t) be rect pulses of amplitude -A or +A. • Mapping to base-band may then be as follows ( ω C =2 π f C ) Bit1 bit2 b R (t) b I (t) QPSK symbol transmitted 0 0 − A − A − Acos( ω C t) − A sin( ω C t) = Acos( ω C t − 135 0 ) 0 1 − A +A − Acos( ω C t) + A sin( ω C t) = Acos( ω C t+135 0 ) 1 0 +A − A Acos( ω C t) − A sin( ω C t) = Acos( ω C t − 45 0 ) Acos( ω C t) + A sin( ω C t) = Acos( ω C t +45 0 ) 1 1 +A +A Looking at a constellation diag for this mapping makes it clear why Acos( ω C t) + A sin( ω C t) = Acos( ω C t +45 0 ) etc. 24 Apr'06 CS3282 Sectn 8 76

Constellation diagram for ± 45 o, ± 135 o QPSK In quadrature Symbol allocation table: with cos 0,1 Bit1 bit2 b R (t) b I (t) 1,1 V 0 0 − A − A In phase 0 1 − A +A with cos 45 o 1 0 +A − A V -V (real pt) 1 1 +A +A 1,0 0,0 24 Apr'06 CS3282 Sectn 8 77

Alternative constellation diag ( 0 o ,90,180,270 o QPSK) Imag pt Symbol allocation table: 0,1 Bit1 bit2 b R (t) b I (t) 0 0 A 0 0,0 1,0 0 1 0 +A 1 0 -A 0 Real 1 1 -A -A 1,1 24 Apr'06 CS3282 Sectn 8 78

QPSK is 4-PSK. What about 8-PSK & 16-PSK? Can have 8-PSK (3 bits/symbol) & 16-PSK (4 bits/symbol). Constellation diagrams for shown below. Imag pt Re Real pt 16- 8-PSK PSK Differential forms of QPSK & M-PSK often used where changes in phase signify the data. Principle similar to DPSK . 24 Apr'06 CS3282 Sectn 8 79

Exercise 8.6: Consider how symbols for 8-PSK & 16-PSK may be associated with sequences of 3 or 4 bits, i.e. label the constellation diagrams. Use a form of 'Gray coding'. 011 001 010 110 000 With Gray 100 coding, a symbol 111 error generally 101 causes just one bit-error 24 Apr'06 CS3282 Sectn 8 80

Exercise 8.6 (cont) : What happens if we don’t use Gray coding? 010 001 011 100 000 111 101 If symbol 111 mistaken for 000 110 get 3 bit-errors 24 Apr'06 CS3282 Sectn 8 81

Advantage of Gray coding •With Gray coding of multi-level symbols, bit-error rate may be assumed to be: symbol-error rate ÷ no. of bits/symbol except when the noise is exceptionally high. (We assume a symbol error just takes us to a nearby symbol which differs in just one bit with Gray coding) • Repeat the labeling now for 16-PSK. 24 Apr'06 CS3282 Sectn 8 82

Exercise 8.7: Show how a vector-modulator may be used to generate the 8 or 16 symbols of 8-PSK & 16-PSK. 011 Symbol b R (t) b I (t) 001 000 V 0 010 V 001 V/1.4 V/1.4 110 000 010 -V/1.4 V/1.4 011 0 V 100 100 V/1.4 -V/1.4 111 101 101 0 -V 110 -V 0 111 -V/1.4 -V/1.4 24 Apr'06 CS3282 Sectn 8 83

Example 8.7 (cont) How would you detect 8-PSK with a vector demodulator & threshold detectors? Exercise 8.8: If radius of constellation diagram circle is V volts for QPSK, 8- PSK & 16-PSK calculate energy per bit for each of these schemes assuming rectangular pulses. Take 'noise immunity' as distance between each symbol on constellation diagram & nearest one to it, Estimate noise immunity for QPSK, 8-PSK & 16-PSK when radius is V in each case. 24 Apr'06 CS3282 Sectn 8 84

Exercise 8.9: How will pulse-shaping be applied to QPSK, 8-PSK and 16-PSK? With 100% RRC pulse shaping & symbol duration T, what is band-with efficiency (in b/s / Hz) for each of these techniques. What is theoretical maximum bandwidth efficiency in each case? 24 Apr'06 CS3282 Sectn 8 85

Single carrier digital modulation schemes •ASK, FSK, PSK, DPSK, QPSK •Differential QPSK •Gaussian FSK & MSK •Combined ASK & PSK (QAM, APK) •etc. 24 Apr'06 CS3282 Sectn 8 86

Other modulation techniques •Direct sequence spread spectrum techniques (DSSS) •Frequency hopping (FHSS) •Complementary code keying (CCK) 24 Apr'06 CS3282 Sectn 8 87

Pause • End of slides on single carrier modulation 24 Apr'06 CS3282 Sectn 8 88

Multi-carrier modulation & OFDM •Orthog frequency division multiplexing (OFDM) is relatively new ‘multi-carrier’ modulation scheme. • Used for DAB, ADSL & wireless LANs (IEEE 802.11a). • Many, say 64 or 1024, carrier frequencies evenly spaced out over a range of frequencies. •Used in IEEE802.11g/a/e with 64 carrier frequencies. •High bandwidth efficiency & robust to freq. selective fading. • First a few preliminaries & reminders 24 Apr'06 CS3282 Sectn 8 89

Quadrature amplitude modulation_QAM • QPSK combined with multi-level ASK • With QPSK, ± A applied to cos & sin carriers • With QAM, 0, ± A, ± 2A applied • Nicely represented by ‘constellations’ 24 Apr'06 CS3282 Sectn 8 90

Constellation for QPSK Modulating sin Bit1 Bit2 b R b I 1,0 0,0 0 0 A A 0 1 A -A modulating 1 0 -A A cos 1 1 -A -A 1,1 0,1 24 Apr'06 CS3282 Sectn 8 91

16-QAM _ symbol allocation table Bit1 bit2 bit3 bit4 VI VQ Bit1 bit2 bit3 bit4 VI VQ 0 0 0 0 A A 1 0 0 0 3A A 0 0 0 1 A -A 1 0 0 1 3A -A 0 0 1 0 A 3A 1 0 1 0 3A 3A 0 0 1 1 A -3A 1 0 1 1 3A -3A 0 1 0 0 -A A 1 1 0 0 -3A A 0 1 0 1 -A -A 1 1 0 1 -3A -A 0 1 1 0 -A 3A 1 1 1 0 -3A 3A 0 1 1 1 -A -3A 1 1 1 1 -3A -3A 24 Apr'06 CS3282 Sectn 8 92

‘16_QAM’ constellation Imag (0010) 3A (0100) (0000) A Real -A A 3A -A (0001) -3A (0011) 24 Apr'06 CS3282 Sectn 8 93

Vector-modulator Sin(2 π f C t) 10110 b I (t) Map Mult ADD Mult Map b R (t) 11011 Cos(2 π f C t) 24 Apr'06 CS3282 Sectn 8 94

Vector modulator in complex notation Take b(t) + jq(t) as a complex b-b signal. cos(2 π f C t).b(t) + sin(2 π f C t). q(t) = real { ( b(t) + jq(t) ) exp(-2 π j f C t) } 10110 b(t) Map Mult Complx 11011 base-band exp(-2 π j f C t) 24 Apr'06 CS3282 Sectn 8 95

A slight variation cos(2 π f C t).b(t) − sin(2 π f C t). q(t) = real { ( b(t) + jq(t) ) exp( 2 π j f C t ) } Instead of sin we modulate -sin: no real difference 10110 b(t) Map Mult Complx 11011 base-band exp(2 π j f C t) 24 Apr'06 CS3282 Sectn 8 96

A slight variation (continued) − Sin(2 π f C t) 10110 b I (t) Map Mult ADD Mult Map b R (t) 11011 Cos(2 π f C t) 24 Apr'06 CS3282 Sectn 8 97

Fast Fourier Transform → {X[k]} 0,N-1 FFT : {x[n]} 0,N-1 − N 1 [ ] [ ] ∑ − π = j 2 kn / N X k x n e for k = 0 , 1 , ..., N- 1 = n 0 |X[k]| x[n] n k 0 N N/2 N 0 f S Time-domain f S /2 0 Frequency domain 24 Apr'06 CS3282 Sectn 8 98

Implementation of FFT The FFT is ‘fast’ in that it can be programmed or implemented in hardware very efficiently especially when N is a power of 2, e.g. 64, 256, 512, 1024, 24 Apr'06 CS3282 Sectn 8 99

Inverse Fast Fourier Transform IFFT: {X[k]} 0,N-1 → {x[n]} 0,N-1 − N 1 1 [ ] ∑ π = j 2 kn / N x n X[k] e for k = 0 , 1 , ...,N- 1 N = n 0 |X[k]| x[n] n k 2N 0 N N/2 N 0 f S Time-domain f S /2 0 24 Apr'06 CS3282 Sectn 8 100