# hw03 - ECE 488 RF Circuits and Systems Spring 2009 HOMEWORK...

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Unformatted text preview: ECE 488: RF Circuits and Systems Spring 2009 HOMEWORK 3: Amplifiers, L-Networks, Mixers, and Filters Reference: Lectures 4–6; Chapters 3–6 Due: February 11 at 11:15 am Problem 1. Do you like puzzles? How about a re- ally tricky engineering design puzzle? This problem requires nothing more than what you have already learned about L-networks, tanks, and basic amplifier operations. The RF power amplifier shown in Fig- ure 1 supplies a maximum power of 2 watts to the 125-ohm load ( R L ). The DC supply voltage is 10 volts as shown. Assume idealized operation of the transistor. [20 points] a. Find the values of X L 2 and X C 2 making up the matching network. Hint: power = V 2 pk / (2 R ) . Warning to TAs: Do not solve this puzzle for the students! TA’s can check with me for further authorized hints. b. List all other hints you obtained and needed to solve this puzzle! 125 R L L 1 L 2 I B +10 V C 2 C 1 Q 1 V L Figure 1: Transistor amplifier. Problem 2: This problem provides information and calculations needed for Lab 2 dealing with mixers and filters. A mixer shifts the carrier fre- quency of a signal to either the sum or the difference frequency between the LO (Local Oscillator) and carrier. Mathematically in the time domain, a balanced switching mixer multiplies the input signal by alternating plus and minus ones cycling at the LO frequency, a square wave. An unbalanced switching mixer mutiplies the input by alternating ones and zeros, again cycling at the LO frequency, thereby creating an offset square wave. The action of these mixers in the time domain effectively “chops up” the input signal. In the frequency domain the switching mixer creates sum and difference frequencies between the carrier and the harmonics of the square waves created by the switches, not just at the LO frequency. In Lab 2 you will build an unbalanced mixer . [20 points] a. For the case of the unbalanced mixer, the switch action can be described over one LO period ( T ) by the function g LO ( t ) = for- 1 2 T < t <- 1 4 T 1 for- 1 4 T < t < + 1 4 T for + 1 4 T < t < + 1 2 T Sketch g LO ( t ) and compute the first several terms of the Fourier series of the function g LO ( t ). Assume a fundamental LO frequency of 1.70 MHz.a fundamental LO frequency of 1....
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hw03 - ECE 488 RF Circuits and Systems Spring 2009 HOMEWORK...

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