hw 7 solutions - E5 31‘ HW 7 Soluh'ons 6.1-2(3 1...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: E5 31‘] HW 7 Soluh'ons 6.1-2 (3) 1 smc(21007rt) 4:) mil (f/2100) The bandwidth of this signal is 2100 1r rad/s or 1050 Hz. The Nyquist rate is 2100 Hz (samples/s). (b) 55inc2 (200m 4=> 0.02m (f/400) The bandwidth of this signal is 400 1r rad/s or 200 Hz. The Nyquist rate is 400 Hz (samples/s). (c) ‘ _ sinc (210mm) + sinc2(2001rt) «l» gang/2100) + 0.005A(f/400) qr." " .. if: in" L _ The bandwidth of the first term on the right—hand side is 1050 Hz, and the bandwidth of the second term is 200 Hz. Clearly the bandwidth of the composite signal is the higher of the two. that is, 1050 Hz. The Nyquist rate is 2100 Hz (samples/s). (d) sinc(2007rt) 4:) 0.005 U (f/200) . l smc (21001rt) <=> m H(f/2100) The two signals have bandwidths 100 Hz, and 1050 Hz, respectively. The spectrum of the product of two signals is 1/27r times the convolution of their spectra. Rom width property of the convolution, the width of the convoluted signal is the sum of the widths of the signals convolved. Therefore, the bandwidth of sinc (2007rt)sinc (21007rt) is 100 + 1050 = 1150 Hz. The Nyquist rate is 2300 Hz. 6.1-8 Assume that a signal g(t) is simultaneously time-limited and band-limited. Let C(f) = 0 for If! > B denote its band-limited characteristics. Therefore G'(f)-H (T27) =G(f) for 3' >8. Therefore from the time-convolution property g(t) = g(t) * 2B’sinc (0nB’t) = 2B’g(t) at sinc (27rB’t) Because g(t) is time-limited, there exists a T such that g(t) = 0 for [t] > T. But 9(6) is equal to convolution of g(t) with sinc (27rB’t) which is not time-limited. The convolution lasts forever. It is impossible to obtain a time~limited signal from the convolution of a time-limited signal with a non-time—limited signal. 6.1-4 I, l (a) 1 x 10(1) = H (Ti — 5) 4:» P(f) = T, sine (fink-infra (b) t 1 T ”T , p“) = H (Ts/2 ' 5) 4:» P(f) = —23sinc ( 2 3) e—wm (C) 1 i pm = sin (27,153) M) — u<t — T./2>1«=> Pm = a 717T: f2 (1+ e—mn) Based on Section 6.1.1, the equalizer filter E(F) should be T,/P(f), lfl S B E(f)= flexible, B<lfl<(1/T,—B) 0» IfIZU/Ts-B) Substituting P( f ) of parts (a) and (b) into the foregoing equation, we can get their corresponding equalizer filter EU). 6.2—2 (a) The Nyquist rate is 2 x 4.5 = 9 MHz. The actual sampling rate is 1.2 x 9 = 10.8 MHz. (b) Since 1024 = 21°, 10 bits or binary pulses are needed to encode each sample. (c) 10.8 x 10 = 108 Mbit/s. The minimal bandwidth is 108/2 = 54 MHz. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern