EECS 216 FALL 2017 HOMEWORK 8 SOLUTIONS

Consider an AM station broadcasting on a carrier frequency of 500 kHz. The radio has a 10 KHz wide fixed IF filter centered at 800 KHz. Find the two possible frequencies for the local oscillator and for each of these two frequencies find the corresponding image frequency. Solution fc = 500 KHz and fIF = 800 KHz fLO1 = |fc − fIF | = |500 − 800| = 300 KHz and fImage1 = |fc − 2fIF | = |500 − 2 × 800| = 1100 KHz fLO2 = fc + fIF = 500 + 800 = 1300 KHz and fImage2 = fc + 2fIF = 500 + 2 × 800 = 2100 KHz End Solution Note: In the same way we defined measures of BW of a signal, we can also define measures of the time duration of a signal. For example, the RMS duration of a signal x(t) is defined as τRMS = sR−∞ ∞ R(−∞ t∞−|tx0()t2)|x|2(dt t)|2dt, where t0 = RR−∞ −∞ ∞∞ t||xx((tt))||22dt dt The following two problems relate to the above definition. 2 Consider the signal x(t) = te−atu(t), a 0. (a) Find the rms time duration of this signal. Hint: Evaluate the required integrals by using the fact that for b 0 dn dbn Z0∞ e−bxdx = db dnn 1b ⇒ Z0∞ xne−bxdx = n!b−(n+1) (b) Sketch |X(jω)|2 and find the 3 dB bandwidth of this signal. (c) Find the rms bandwidth of the signal. Hint: Evaluate the needed integral by making a trigonometric substitution ω = a tan θ. Solution (a) t0 = R0∞ tx2(t)dt R0∞ x2(t)dt = 6(2a)−4 2(2a)−3 = 3 2a Z0∞(t − t0)2x2(t)dt = Z0∞ t2x2(t)dt − 2t0 Z0∞ tx2(t)dt + t2 0 Z0∞ x2(t)dt = 4!(2a)−5 − 2( 3 2a )3!(2a)−4 + ( 3 2a )22!(2a)−3 1 EECS 216 FALL 2017 HOMEWORK 8 SOLUTIONSThus τ 2 rms = R0∞(t − t0)2x2(t)dt R0∞ x2(t)dt = 4!(2a)−5 − 2(23a)3!(2a)−4 + (23a)22!(2a)−3 2(2a)−3 = 3 4a2 and so τrms = √3 2 1 a (b) X(jω) = 1 (a + jω)2 Therefore |X(jω)|2 = 1 (a2 + ω2)2 Thus 1 2 1 4a = 1 (a2 + ω32dB)2 → ω3dB = [√2 − 1]1/2a = 0.644a (c) ω2 rms = R0∞ ω2|X(jω)|2dω R0∞ |X(jω)|2dω = R0∞ (a2+ω2ω2)2 dω R0∞ (a2+1ω2)2 dω Note: By making the trigonometric substitution ω = a tan θ, one gets Z0∞ (a2 +ω2ω2)2 dω = a1 Z0π/2 sin2 θdθ = 21a Z0π/2(1 − cos 2θ)dθ = 4πa Z0∞ (a2 +1ω2)2 dω = a13 Z0π/2 cos2 θdθ = 21a3 Z0π/2(1 + cos 2θ)dθ = 4πa3 Thus ωrms = a End Solution 3 Consider the signal p(t) given by p(t) = A exp(−τt22 ) where A and τ are constants. 2(a) Use the following integrals to find the rms time duration of the signal this baseband signal. Z0∞ x2 exp(−ax2)dx = 41arπa Z0∞ exp(−ax2)dx = 1 2rπa (b) Find P(jω). Find the rms bandwidth of the signal x(t). Hint: exp(−at2), a 0 F.T. ↔ rπa exp(−ω4a2 ) are a Fourier transform pair. Solution (a) By symmetry t0 = R−∞ ∞ tp2(t)dt R−∞ ∞ p2(t)dt = 0 and thus τ2 rms = R−∞ ∞ (t − t0)2p2(t)dt R−∞ ∞ p2(t)dt = R0∞ t2e−2t2/τ2dt R0∞ e−2t2/τ2dt = τ2 4 → τrms = τ 2 (b) exp(−at2), a 0 F.T. ↔ pπa exp(−4ωa2 ) and thus Ae−t2/τ2 F.T. ↔ Aτ√πe−ω2τ2/4 Therefore ω2 rms = R0∞ ω2|P(jω)|2dω R0∞ |P(jω)|2dω = R0∞ ω2e−ω2τ2/2dω R0∞ e−ω2τ2/2dω = 1 2τ → ωrms = 1 τ (c) From parts (a) and (b) above ωrmsτrms = 1 τ τ 2 = 1 2 Note that for any signal, τrmsωrms ≥ 1/2 (this is the celebrated “uncertainty principle”), and thus the signal given in this problem achieves the minimum possible value for the rms time-bandwidth product