ENGR43 Test 2 Spring 29 ENGR43 Spring 29 Test 2 Name: SOLUTION Section: 1(MR 8:) 2(TF 2:) 3(MR 6:) (circle one) Question I (2 points): Question II (2 points): Question III (17 points): Question IV (2 points): Question V (23 points): Total (1 points): On all questions: SHOW ALL WORK. BEGIN WITH FORMULAS, THEN SUBSTITUTE VALUES AND UNITS. No credit will be given for numbers that appear without justification. 1 of 13
ENGR43 Test 2 Spring 29 Question I Bridges and Damped Sinusoids (2 points) 6 6 4 2 V( 2 4 6 6 4 2.5. 1 5. 1 4 7.5. 1 4 ms t.1.125.15.175.2 2ms 1) (4 pts) Find the period and frequency of this signal. Include units. T=1ms/3 cycles they can also see that ~.33 ms for one cycle or 2ms/6 cycles =.33333 s f=1/t = 3k 2) (2 pts) What is the value of the DC offset? Offset= 2V 3) (4 pts) Write the mathematical expression for this signal and its offset. In general, this is given by X= X X 1 sin (ωt Φ ) which accounts for its frequency, phase shift, and offset. (Phase shift with respect to t=) Phase shift = 9 or π/2 2 3V sin (2 * π * 3k * t π/2) = 2V 3V sin (18,85t π/2) 2 of 13
ENGR43 Test 2 Spring 29 The circuit below gives the simulation output below. TOPEN = 1 2 U1 R1 5 L1 1 2 3mH V 1V V1 1 2 U2 TCLOSE = C1.5uF 4) (3 pts) Find the damping constant α for this data (use points labeled for you to three significant figures). (t, V ) = (ms, 1V) (t 1, V 1 ) = (2.68 ms, 1.7V) V 1 =V e α(t 1 t ) α=834 1/s 3 of 13
ENGR43 Test 2 Spring 29 5) (4 pts) Write the mathematical expression for this data in the form of v(=a cos (ω e αt f=12 cycles/3ms=4k Hz or they can find the period and inverse ω=2*π*f = 25,132 rad/s or 25 K rad/s 1V cos (25 k * e 834t 6) (3 p If the resistor is completely removed from the circuit and replaced with a wire, what will happen to the frequency and amplitude of the simulated output? Why? The frequency will remain the same. The amplitude will no longer be dampened and it will forever oscillate sinusoidally. The frequency is not affected by the resistor. The amplitude no longer reduces with time because there is nothing to slow the transfer of energy between the potential energy and kinetic energy from the capacitor and inductor respectively. (conservation of energy) 4 of 13
ENGR43 Test 2 Spring 29 Question II Thevenin Equivalents (2 points) R1 2k A 1V V2 4k R4 2k RL B R5 1) (6 pts) Find the Thevenin voltage (Vth) of the circuit inside the rectangle. R 245 := R 2 R 4 R 5 R 245 = 4 1 3 Ω R 3 R 245 R 2345 := R 3 R 245 R 2345 = 2 1 3 Ω 1V 2kΩ V 2345 := 4kΩ V 2345 = V 4 V A V B 5 V V th R 4 V 4 := V 2345 R 2 R 4 R 5 V 4 = 2.5 V V th := 2.5V 2) (6 pts) Find the Thevenin resistance. Short the voltage supply and if it helps turn the circuit around A R 1 R 3 R 13 := R 1 R 3 R 13 = 1.333 1 3 Ω B R4 2k 4k R1 2k R 1325 := R 2 R 5 R 13 R 1325 = 3.333 1 3 Ω R5 R 4 R 1325 R 41325 := R 4 R 1325 5 of 13 R 41325 = 1.25kΩ R th 1.25kΩ
ENGR43 Test 2 Spring 29 3) (4 pts) Draw the Thevenin equivalent circuit with the load (1K ohms). Rth 1.25k 2.5V Vth RL 4) (4 pts) For the following Thevenin equivalent circuit (not the one above) find the voltage between A and B with a load of 2K ohms. Rth A 3V Vth RL 2k B 6 of 13
ENGR43 Test 2 Spring 29 Question III OpAmp Applications (17 points) U1.5Vac Vdc Vin Va U2 R1 Vb R5 C1 R8 U3 U4 Vc R6 U5 Vout R9 R4 Vd R7 C2 Assume that ±9 Volt power supplies have been properly connected to all five opamps in the circuit above. 1) (5p The circuit has 5 opamps labeled as U1 through U5. State what the opamp circuit is for each. Choices are: 1. Follower/Buffer, 2. Inverting Amp, 3. Noninverting Amp, 4. Differentiator, 5. Integrator, 6. Adding (Mixing) Amp, 7. Difference (Differential) Amp, 8. Miller Integrator, 9. Practical Differentiator. U1 Circuit: _FOLLOWER_ U2 Circuit: _NONINVERTING_ U3 Circuit: _DIFFERENTIATOR_ U4 Circuit: _INTEGRATOR_ U5 Circuit: _ADDING 2) (4p Determine the values relative to ground, of Va(, Vb(, Vc( and Vd( as functions of Vin( with R1 = 2k, = 2k, = 4k, R4 = 3k, R5 =, R6 = 2k, R7 = 3k, R8 =, R9 =, C1 = 1µF and C2 = 68µF. a) Voltage at point Va(: Va( = Vin( Follower output = input b) Voltage at point Vb(: R1 Vb = 1 Va( = ( 2k ( 1 2k ) Vin( = 2Vin( t ) 7 of 13
ENGR43 Test 2 Spring 29 c) Voltage at point Vc(: dva( dvin( Vc( = µ dt dt ( C1) = ( 4k 1 ) = (.4) dvin( dt d) Voltage at point Vd(: 1 1 Vd ( = Va( dt= Vin( dt= 4.92 Vin( dt R4 C2 3k 68µ 3) (3p Determine the output voltage, Vout(, as a function of Vb(, Vc( and Vd(. R8 R8 R8 Vout( = Vb( Vc( Vd( R5 R6 R7 = [ Vb( ] Vc( ] [ Vd( ] 2k 3k = Vb(.5Vc(.333Vd( 4) (4p Now find Vout( as a function of Vin(. Vout( = Vb(.5Vc(.333Vd( dvin( = [2Vin( ].5[.4 ].333[ 4.92 dt dvin( = 2Vin(.2 1.634 Vin( dt dt 5) (1p What is a practical use for a voltage follower/buffer opamp? Vin( dt] 1. Prevent a load or the circuit being driven from changing the characteristics of the previous stage 2. Provide a high input impedance for a circuit s output 3. Provide a power gain but not a voltage gain 8 of 13
ENGR43 Test 2 Spring 29 Question IV OpAmp Analysis (2 points) V1 R1 V2 Vout R4 Assume that ±9 Volt power supplies have been properly connected to the opamp in the circuit above. 1) (2p What opamp circuit given on your crib sheet does this circuit most closely represent? Adder (Mixing) Amplifier 2) (3p What feedback connection is necessary before the golden rule concerning the voltages on the two input terminals can be applied? The opamp must have negative feedback from its output to the negative input terminal. This can be a plain wire or a resistor. 3) (1p Using the two golden rules of opamp analysis, derive an expression for Vout in terms of V2 and the appropriate resistor values, when V1 has been set to Volts. You must use Ohm s Law and current laws to describe how the opamp functions by the golden rules. V = V = V since V is grounded No current will flow into V and since V1 = V, no current will flow through R1 ( V=). Therefore all the current through must go through. V 2 Vout = Vout = V 2 9 of 13
ENGR43 Test 2 Spring 29 4) (4p Remove R4, set V2 = V1 (short the 2 inputs together), and derive the expression for Vout in terms of V1 and the resistors using the rules for opamp analysis. V = V = V R12 = R1 = R1/(R1) The current through R12 equals the current through (no current into the opamp V inpu V1 Vout = R12 R1 Vout = V1= V1 R12 R1 5) (1p What does R4 model or represent in this circuit? R4 represent the load resistor on the opamp or the load being driven by the opamp such as a motor, speaker, or LED 1 of 13
ENGR43 Test 2 Spring 29 Question V OpAmp Integrators and Differentiators (23 points) In the circuit below, Rin = 2Ω, Lf = 2mH and Rf = 2Ω. Assume that power supplies have been properly connected to an ideal opamp and the output voltage is limited to ±1V Zf V1 Rin Lf 1 2 1) (4p Find the transfer function H(jω) = Vout(jω)/V1(jω) for this circuit. (Substitute in the values provided for the components.) V Zf jωlf R out f jω.2 2 2 j(.2) ω H ( jω) = = = = = 1 j. 1ω V Z R 2 2 1 in in Rf Rload 2) (2p What function is this circuit designed to perform at high frequencies (f 2kHz)? V jωl jω.2 ω = ω ω V R 2 out f H ( j ) = = = j.1 for >> 1 in R L f f Differentiation (with gain of.1) 3) (2p What function is this circuit designed to perform at low frequencies (f 1Hz)? V R 2 ω = ω V R 2 out f H ( j ) = = = 1 for << 1 in R L f f Inverting amplifier (with gain of 1) 4) (4p What is the corner frequency (in Hz) for the circuit where it transitions from its low frequency performance to its high frequency performance? ω c => R f = jωl f ω c = R f /L f = 2/.2 = rad f c = ω c /2π = 159.2Hz 11 of 13
ENGR43 Test 2 Spring 29 5) (4p Sketch the bode plot of H(jω), 2log(magnitude) vs. log of frequency, of the transfer function from.1 to Hz. For simplicity assume that the ideal opamp output voltage is unlimited. 4 2 1mHz 1.Hz 1Hz 1Hz 1.KHz 1KHz DB(V(Rin:2)) Frequency 6) (4p For the input below, sketch the output on the same axis. Note the frequency of the input signal. 1V 5V V 5V 1V s.4ms.8ms 1.2ms 1.6ms 2.ms V(Rin:1) Time 7) (2p For the input below, sketch the output on the same axis. Note the frequency of the input signal. 1V 5V V 5V 1V s.2s.4s.6s.8s 1.s V(Rin:1) Time 12 of 13
ENGR43 Test 2 Spring 29 4 2 1mHz 1.Hz 1Hz 1Hz 1.KHz 1KHz DB(V(Rload:2)) 159.2Hz 1V Frequency 5V V 5V 1V s.4ms.8ms 1.2ms 1.6ms 2.ms V(Rin:1) V(Rload:2)/1 Time 1V 5V V 5V 1V s.2s.4s.6s.8s 1.s V(Rin:1) V(Rload:2) Time 13 of 13