Solns ch3 Upamanyu Madhow PDF

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Upamanyu Madhow Solutions Chapter 3...

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Solutions to Chapter 3 Problems Introduction to Communication Systems, by Upamanyu Madhow Problem 3.1(a) The AM signal is given by u(t) = (Ac + m(t)) cos 2πfc t where m(t) = A cos 2πfm t. The largest negative excursion of this message has magnitude A, so that the modulation index is amod = A/Ac. The largest value of u(t) is Ac + A = 30, and the smallest value is Ac − A = 10, so that Ac = 20, A = 10, and the modulation index is 0.5. (b) The power is given by u2 = (Ac + m(t))2 cos2 2πfct = (Ac2 + m2 (t) + 2Ac m(t))(1 + cos 4πfc t)/2 = Ac2/2 + m2 /2 + Ac m + 12 (Ac2 + m2 (t) + 2Ac m(t)) cos 4πfct) The term involving cos 4πfc t is passband around 2fc , and hence has zero DC value. We also have m2 = A2 / and m = 0 for our sinusoidal message, so that the power is given by Pu = u2 = (Ac2 + A2 )/2 = 250 (c) We see from the figure that the envelope has period 0.5 ms, i.e., our sinusoidal message has frequency 2 KHz. Thus, the bandwidth of the AM signal is 4 KHz. M(f) exp(−j

/4)

4 exp(−j

/4) 4 exp(j /4)

U(f) 4 exp(−j

/4) 4 exp(j /4)

exp(j /4) f

−201

−199

199

f

201

Figure 1: Spectra of message and DSB signal (Problem 3.2).

V p(f) 4 exp(−j /4)

4 exp(j /4)

−201

201

USB signal

f

V(f)

8 exp(j /4)

1

f

Complex envelope

Figure 2: Spectra of the USB signal and its complex envelope (Problem 3.2).

Problem 3.2(a) We have π π π π π m(t) = 2 cos(2πt + ) = ej(2πt+ 4 ) + e−j(2πt+ 4 ) ↔ M(f ) = ej 4 δ(f − 1) + e−j 4 δ(f + 1) 4

For the DSB signal, we have u(t) = 8m(t) cos 400πt = 4m(t)ej400πt + 4m(t)e−j400πt ↔ U (f ) = 4M(f − 200) + 4M(f + 200) which is sketched in Figure 1. (b) The output of the envelope detector is 8|m(t)|, which is a rectified version of the message (illustrating that envelope detection does not work for extracting the message from a DSB signal). Plot omitted.

(c) The spectrum of the passband signal and its complex envelope with respect to 200 Hz are sketched in Figure 2. Thus, the complex envelope is given by π

π

V (f ) = 8ej 4 δ(f − 1) ↔ v (t) = 8ej(2πt+ 4 ) Taking the real and imaginary parts of v(t), we obtain the I and Q components of the USB signal π vs (t) = 8 sin(2πt + ) 4

π vc (t) = 8 cos(2πt + ), 4

(We decided to do this from scratch, but we could have used the expression for a USB signal in terms of the message and its Hilbert transform given in the text.) |M(f)|

2 3/2 −3

−1

1

3

f(KHz)

Figure 3: Message magnitude spectrum for Problem 3.3(a).

|UAM(f)|

1

0.113 597

599 600 601

0.151 603

f (KHz)

Figure 4: Magnitude spectrum of AM signal in Problem 3.3(e). Only positive frequencies are shown, since the magnitude spectrum is symmetric. Problem 3.3 (a) The message consists of tones at 1 KHz and 3 KHz, hence message bandwidth is 3KHz. We have j 2πt

−j 2πt

j6πt

+ 4 e −e m(t) = 3 cos 2πt + 4 sin 6πt = 3 e +e 2j 2 ↔ M(f ) = 23 (δ (f − 1) + δ (f + 1)) + j2 (δ (f − 3) − δ (f + 3)) −j6πt

The magnitude spectrum is sketched in Figure 3. (b) We have −M0 = minx 3 cos x + 4 sin 3x. Numerical minimization in Matlab (need only to do it over a period) yields M0 = 6.627, so that mn (t) = m(t)/M0 = 0.4527 cos 2πt + 0.6036 sin 6πt. (c) Time domain expression for AM signal with modulation index 50% and carrier frequency 600 KHz is given by uAM (t) = Ac(1 + amod mn (t)) cos 2πfct = Ac (1 + 0.226 cos 2πt + 0.302 sin 6πt) cos 1200πt (d) Noting that mn2 = 0.45272/2 + 0.60362/2 = 0.2846, the power efficiency is given by ηAM =

2 m2n amod = 0.066 2 m2 1 + amod n

or 6.6%. (e) Taking Ac = 2 for convenience, the spectrum of the AM signal is given by UAM (f ) = δ(f − fc ) + δ(f + fc ) + amod (Mn (f − fc ) + Mn (f + fc )) 2

where Mn (f ) = 0.226 (δ(f − 1) + δ(f + 1)) + 0.302 (δ (f − 3) − δ (f + 3)). Since |UAM (−f )| = j |UAM (f )| (magnitude spectrum for real-valued uAM (t) must be symmetric), it suffices to specify the spectrum for positive frequencies as |UAM (f )| = δ(f −fc )+amod|Mn (f −fc )| = δ(f −600)+0.113 (δ(f − 601) + δ(f − 599))+0.151 (δ(f − 603) − δ(f setting fc = 600 and amod = 0.5. The magnitude spectrum is sketched in Figure 4. (f) We need the RC time constant to be large enough to average out carrier cycles, but small enough to track message variations: 1 1 ≪ RC ≪ fc B For R = 50 ohms, fc = 600 KHz and B = 3 KHz, we get that 3.3 × 10−8 ≪ C ≪ 6.6 × 10−6 farads For example, C = 500 nF would work. (A cos

Re(UUSB(f))

)/4

(A cos

)/4

fc + f m

−fc − m f

(A sin

Im(UUSB(f))

)/4

−fc − m f fc + f m −(A sin

)/4

(A cos

)/4

Re(ULSB(f))

(A cos

−fc + fm (A sin

)/4

f c− fm )/4

Im(ULSB (f)) f c− fm

−fc + fm −(A sin

)/4

Figure 5: SSB spectra for Problem 3.4(a). Problem 3.4 The message is given by 1 ejφ e−jφ 1 m(t) = cos(2πfm t + φ) = ej(2πfm t+φ) + e−j(2πfm t+φ) ↔ M(f ) = δ(f − fm ) + δ(f + fm ) 2 2 2 2 and the DSB signal is given by up (t) = Am(t) cos 2πfct ↔ Up (f ) = 2AM(f − fc ) + A M(f + fc ) 2 jφ −jφ jφ = Ae4 δ(f − fc − fm ) + Ae4 δ(f − fc + fm ) + + Ae4 δ(f + fc − fm ) +

Ae−jφ δ(f 4

+ fc + fm )

(a) The USB signal is given by UUSB (f ) =

Aejφ Ae−jφ δ(f + fc + fm ) δ(f − fc − fm ) + 4 4

3

(1)

and the LSB signal is given by ULSB (f ) =

Aejφ δ(f − fc + fm ) +

Ae−jφ δ(f + fc − fm ) 4

(2)

4 with spectra sketched in Figure 5. (b) Taking inverse Fourier transforms in (1) and (2), we obtain that the time domain expressions for the SSB signals are given by uU SB (t) =

A cos (2π(fc + fm ) + φ) 2

A cos (2π(fc − fm ) + φ) 2 Alternatively, one can find the complex envelopes of these signals in the frequency domain and jφ then go to the time domain. For example, the complex envelope of the USB signal is Ae2 δ(f − jφ fm ) ↔ u˜(t) = Ae2 ej2πfm t , so that the passband signal is     A A j(2π (fc +fm )t+φ) j2πfc t e = cos (2π(fc + fm ) + φ) uU SB (t) = Re u˜(t)e = Re 2 2 uLSB (t) =

Yet another approach is to use the Hilbert transform: uU SB (t) =

A A m(t) cos 2πfct − m(t ˇ ) sin 2πfct 2 2

and note that m ˇ (t) = cos(2πfm t + φ − 2pi) = sin(2πfm t + φ), so that uU SB (t) =

A A A cos(2πfm t + φ) cos 2πfct − sin(2πfm t + φ) sin 2πfct = cos (2π(fc + fm ) + φ) 2 2 2

(the computations for

Figure 6: Spectrum for Problem 3.5. Problem 3.5 For x(t) = m(t) + α cos 2πfct, the output of the nonlinearity is given by 2

y(t) = βx2 (t) + x(t) = β (m(t) + α cos 2πfc t) + m(t) + α cos 2πfc t = βm2 (t) + 2βαm(t) cos 2πfct + βα 2 cos2 2πfc t + m(t) + α cos 2πfc t 2 2 = α (1 + 2βm(t)) cos 2πfc t + βm2 (t) + β 2α + β α2 cos 4πfct with the first term being the desired AM waveform. Taking Fourier transforms, we obtain Y (f ) = α2 (δ (f − fc ) + δ(f + fc )) + β (M(f − fc ) + M(f + fc )) 2 2 + β(M ∗ M)(f ) + β α2 δ(f ) + β α4 (δ (f − 2fc ) + δ(f + 2fc)) 4

The first line is the desired AM term around fc (spanning 890 to 910 KHz). The second line corresponds to baseband terms (the convolution M ∗ M has bandwidth 20 KHz, twice that of M , and there is a DC term corresponding to the delta function at the origin) and a sinusoid around 2fc (1800 KHz). The magnitude spectrum is sketched in Figure 6. (b) As shown in Figure 6, the desired AM signal lies in a 20 KHz band around 900 KHz. We can use a BPF around 900 KHz with fairly relaxed bandwidth specs (e.g., a bandwidth of 50 KHz), and comfortably reject both the undesired terms at DC (which goes from 0 to 20 KHz) and that at 1800 KHz. |UDSB(f)| 5/2

5/2

−101

−99

99

101

f (KHz)

Figure 7: DSB spectrum for Problem 3.6(a).

U LSB(f) 5/2

5/2

−100 −99

f (KHz)

99 100

Complex envelope with respect to f c = 100 KHz

5

−1

0

Figure 8: LSB spectrum and its complex envelope for fc = 100 KHz for Problem 3.6(b).

UVSB(f) 5/2

99.5 100.5 101

f (KHz)

Complex envelope with respect to f c = 100 KHz U(f)

A(f)

5

−0.5

0.5 1

f (KHz)

B(f)

+

5

= −0.5

0.5 1

f (KHz)

5

0.5 1

f (KHz)

Figure 9: VSB spectrum and its complex envelope for fc = 100 KHz for Problem 3.6(c)-(d). Problem 3.6 (a) We have m(t) = sinc2t ↔ M(f ) = 21I[−1,1](f ), so that the message bandwidth is 1 KHz. The DSB signal uDSB (t) = 10m(t) cos 2πfc t ↔ 5 (M(f − fc ) + M(f + fc )), where fc = 100 KHz. The spectrum is sketched in Figure 7. (b) The LSB spectrum and its complex envelope with respect to the original carrier frequency fc = 100 KHz are sketched in Figure 8. For this simple scenario, the inverse Fourier transform of the LSB spectrum immediately gives us uLSB (t) = 25 sinct cos 199πt. However, suppose that we wish to express it as a passband signal with reference frequency equal to the original carrier frequency fc = 100 KHz. We can take the the inverse Fourier transform of the corresponding complex envelope to obtain the time domain complex envelope sincte−jπt = sinct cos πt − jsinct sin πt. We can read off the I and Q components as the real and imaginary parts, respectively, and conclude that the corresponding passband signal is uLSB (t) = sinct cos πt cos 200πt + sinct sin πt sin 200πt

5

(c) We sketch the magnitude spectrum, which is symmetric, only for positive frequencies (the spectrum is actually real-valued, and hence equals its magnitude spectrum) in Figure 9. The corresponding complex envelope for fc = 100 KHz, which is used in part (d), is also sketched. (d) We can now compute the inverse Fourier transform of the complex envelope U (f ) in (c) and read off the I and Q components as the real and imaginary parts. Alternatively, we can compute the I and Q components in the frequency domain and then take inverse Fourier transforms. Let’s U (f )+U ∗ (−f ) = 25 I[−1,1](f ) ↔ uc(t) = 5sinc2t, do a bit of both: we see from Figure 9 that Uc (f ) = 2 which is a scaled version of the message, as it should be for a well-designed VSB filter. To find the Q component, let us take the inverse Fourier transform of the complex envelope and read off the imaginary part. Before doing the inverse Fourier transform, let us break U (f ) into two pieces, A(f ) and B(f ), as shown in Figure 9. We have B(f ) = 5I[0.5,1](f ) ↔ b(t) = 25sinc 2t ej1.5πt . To find a(t), first differentiate in the frequency domain to get d A(f ) = 5I[−0.5,0.5](f ) − 5δ (f − 0.5) ↔ 5sinct − 5ejπt df Since dfd A(f ) ↔ −j2πta(t), we obtain a(t) = u(t) = a(t) + b(t) =

5sinct−5ejπt . −j2πt

We therefore have

t 5sinct − 5ejπt 5 + sinc ej1.5πt 2 2 −j2πt

We should check that taking the real part gives, upon simplification, the same answer as before for uc(t). Taking the imaginary part, we get us (t) =

t 5sinct − 5 cos πt 5 + sinc sin 1.5πt 2 2πt 2

Problem 3.7: (a), (b) Instead of considering a term at fL and fH as suggested, it suffices to consider a sinusoid at a generic frequency f ∈ [fL , fH ]: m(t) = cos(2πf t + φ). This is because any real-valued message occupying the band [fL , fH ] can be expressed as a linear combination of real-valued sinusoids: Z fH |M (f )| cos (2πf t + ∠M (f )) m ( t) = fL

Since the processing in Weaver’s modulator is linear, it suffices to check that it works for a generic sinusoid in the band of interest. For m(t) = cos(2πf t + φ), the first set of mixers lead to a sum frequency f + f1 and a difference frequency |f − f1 |. Since fL ≤ f ≤ fH and f1 = (fL + fH )/2, the sum frequency is bounded as fL + f1 = (3fL + fH )/2 ≤ f + f1 ≤ (fL + 3fH )/2 The sum frequency is therefore rejected by the LPF, whose bandwidth W = (2fL + fH )/2 is smaller than the lower bound above. The difference frequency satisfies |f − f1 | ≤ (fL + fH )/2, and is therefore passed by the LPF. Using trigonometric identities, cos A cos B =

1 1 1 1 cos(A − B) + cos(A + B), cos A sin B = − sin(A − B) + sin(A + B ) 2 2 2 2

(here A = 2πf t + φ and B = 2πf1 t). The difference terms passed by the LPF are therefore given by a(t) = 21 cos (2π(f − f1 )t + φ) on the upper branch, and b(t) = −21 sin (2π(f − f1 )t + φ) on the lower branch. (c), (d) The passband output is given by up (t) = a(t) cos 2πf2 t ± b(t) sin 2πf2 t = 21 cos (2π(f − f1 )t + φ) cos 2πf2 t ∓ 21 sin (2π(f − f1 )t + φ) sin 2π = 12 cos (2πf2 t ± (2π(f − f1 )t + φ)) 6

Figure 10: Spectrum of filter output in Problem 3.8(b).

Adding the two branches corresponds to a USB signal with carrier frequency f2 − f1 . Subtracting the two branches corresponds to an LSB signal with carrier frequency f2 + f1 . Problem 3.8 (a) We have fc = 10.7 MHz (the IF frequency), and message bandwidth B = fm = 1 MHz. The RC time constant should satisfy 106 = B ≪

1 ≪ fc = 1.07 × 107 RC

For R = 100Ω, we obtain

1 × 10−9 ≪ C ≪ 10−8 1.07 This is satisfied, for example, by C = 3 × 10−9 F, or C = 3 nF. (b) As shown in Figure 10, the lower sideband is attenuated (slightly) by the filter relative to the upper sideband, so the filter output may be thought of as a VSB signal. The complex envelope is given by 4 6 y(t) = e−j2πfm t + 5 + ej2πfm t 20 20 We get the I and Q components by taking the real and imaginary parts of this: yc(t) = 5 + ys (t) =

1 cos 2πfm t 2

1 sin 2πfm t 10

DSB spectrum 5

−152 −150 −148

148

150

152

f

Figure 11: DSB spectrum for Problem 3.9(a).

7

USB spectrum 5

−152 −150

150

f

152

Complex envelope 10

0

f

2

Figure 12: USB spectrum and complex envelope for Problem 3.9(d).

VSB spectrum 5

149

151 152

Complex envelope

j Us (f)

Uc (f) 5

10

−1

f

0

2

f

−2

0

5

2

f

−2 −1 0 1 2

f

−5

Figure 13: VSB spectrum, complex envelope, and I/Q components for Problem 3.9(e).

8

Problem 3.9(a) The DSB spectrum is given by UDSB (f ) = 5M(f − 150) + 5M(f + 150), and is shown in Figure 11. Bandwidth is 4 Hz (our default assumption is that the unit of frequency is Hz, unless specified otherwise). The message is finite energy, and hence so is the DSB signal. Thus, the power is zero. To compute energy, note that u2DSB (t) = 100m2 (t) cos2 300πt = 50m2 (t) + 50m2 (t) cos 600πt The second term on the extreme right hand side is passband, and hence integrates to zero. Thus, the energy of the DSB signal is given by Z Z 2 EuDSB = 50Em = 50 |M(f )|2 df = 50 12 df = 200 −2

using Parseval’s identity. (b) We have m(t) = 4sinc4t. The output of the envelope detector is 10|m(t)| = 40|sinc4t|. Sketch omitted, since the shape is well-known. The envelope is the magnitude of the message, so sign information is lost. (c) The AM signal is uAM (t) = (A + 4sinc4t) cos 300πt. Thus, we must have A > 4|mint sinc4t| = 4|minx sincx|. The minimum of the sinc function is near the first negative lobe, around x = 1.5. It is actually achieved at a value of x a little less than 1.5, but let’s just approximate the minimum as the value at x = 1.5: minx sincx ≈ sin1.1.5π = − 3π2 = −0.2122. (The actual minimum can be 5π numerically computed as -0.2172.) In particular, A ≥ 0.88 will work. (d) The USB signal has spectrum and complex envelope shown in Figure 12. Taking the inverse Fourier transform of the complex envelope (with respect to fc = 150), we obtain u(t) = 20sinc2tej2πt Taking real and imaginary parts, we obtain: uc(t) = 20sinc2t cos 2πt us (t) = 20sinc2t sin 2πt We see that uc(t) = 20

sin 2πt sin 4πt = 20sinc4t = 5m(t) cos 2πt = 10 2πt 2πt

We can also check that us (t) = 5m ˇ ( t) (e) Filtering the DSB spectrum in (a) using the VSB filter, we obtain the spectrum (only positive frequencies shown) and complex envelope as shown in Figure 13. Also shown are the I and Q components in the frequency domain, obtained in terms of the complex envelope as follows: Uc (f ) = (U (f ) + U ∗ (−f ))/2,

jUs (f ) = (U (f ) − U ∗ (−f ))/2

Clearly, Uc (f ) = 5M(f ), so uc (t) = 5m(t) = 20sinc4t. Finding the Q component in the time domain is an exercise in taking inverse Fourier transforms that we skip here, since similar computations in done in Problem 3.10(f). Problem 3.10(a)-(c) Since hp (t) is real-valued, we have Hp (−f ) = Hp∗(f ). The passband transfer function is sketched in Figure 14. (b) The complex envelope H(f ) = 2Hp+(f + fc ) for fc = 100 is sketched in Figure 14. (c) The I and Q components are given in the frequency domain as Hc(f ) = (H(f ) + H ∗ (−f ))/2 and Hs (f ) = (H(f ) − H ∗ (−f ))/(2j). Since H(f ) is real-valued here, we have Hc(f ) is real and Hs is imaginary, with Re (Hc (f )) = (H(f ) + H(−f ))/2 , Im (Hc(f )) = 0 Re (Hs (f )) = 0 , Im (Hc (f )) = (H(−f ) − H(f ))/2 9

Figure 14: Frequency domain filter characterization for Problem 3.10.

Figure 15: Message spectrum and DSB spectrum for Problem 3.10. Both signals are real-valued in time domain, and hence conjugate symmetric in frequency domain. We only show positive frequencies for the DSB spectrum.

10

Figure 16: Spectrum of Q component of VSB signal in Problem 3.10.

Figure 17: Using differentiation to find the time domain expression for the Q component of the VSB signal in Problem 3.10.

11

The results are sketched in Figure 14. (d) The message signal is given by m(t) = 4sinc4t − 2 cos 2πt ↔ M(f ) = I[−2,2] (f ) − δ(f − 1) − δ(f + 1) The DSB spectrum (ignoring constant factors) is given by UDSB (f ) = M(f − fc ) + M(f + fc ). The message spectrum and the DSB spectrum are sketched in Figure 15. (e) The complex envelope of the VSB signal is given by y = m∗h = m∗(hc +jhs ) = m∗hc +jm ∗hs , so that the I component is given by yc = m ∗ hc and the Q component by ys = m ∗ hs . Going to the frequency domain, the I and Q components are given by Yc(f ) = M(f )Hc (f ) and Ys (f ) = M(f )Hs (f ). From Figure 14, we see that Hc (f ) is flat over [−2, 2], the message band, so that Yc(f ) = M(f ). This is already sketched in Figure 15. For the Q component, we multiply Hs (f ) in Figure 14 and M(f ) in Figure 15 to obtain the spectrum shown in Figure 16. (f) From Figure 16, we see that the Q component can be written as Ys (f ) = jA(f ) + j (δ (f − 1) − δ (f + 1)) where A(f ) is as shown in Figure 17. The inverse Fourier transform of the impulsive term is a sinusoid:   j (δ (f − 1) − δ (f + 1)) ↔ j ej2πt − e−j2πt = j × 2j sin 2πt = −2 sin 2πt

To find the inverse Fourier transform of the non-impulsive term (i.e., to find ja(t)) we employ differentiation. B(f ) = dA(f )/df ↔ b(t) = (−j2πt)a(t) From the graph of B(f ) in Figure 17, we can immediately write down its inverse Fourier transform: b(t) = e−j4πt + ej4πt − 2sinc2t = 2 cos 4πt − 2sinc2t Thus,

b( t) b(t) sinc2t − cos 4πt =− = πt −j2πt 2πt The time domai...