Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

Document related concepts

no text concepts found

Transcript

For this potential divider circuit: RA Vin Vout (input) RB (output) a) Express Vout in terms of Vin, RA and RB b) If Vin is a dc voltage of 6V, what is Vout? c) If Vin is a voltage source which varies with time according to the equation Vin(t) = 20 sin (2π × 50 t), what is Vout? d) If RA = RB, express the ratio Vout / Vin in decibels. • This problem is designed to show that potential dividers behave the same with dc and ac signals, and the concept of gain expressed in decibels. • (a) The output voltage is given by the usual formula: ⎛ RB ⎞ Vout = Vin × ⎜ ⎟ ⎝ RA + RB ⎠ ELEC166 Worked Examples Week 5 1 • (b) With Vin = 6V, the output is obviously ⎛ RB ⎞ Vout = 6 × ⎜ ⎟ ⎝ RA + RB ⎠ , that is, we just RB multiply Vin by the ratio RA + RB . • (c) If Vin(t) = 20 sin (2π × 50 t), we do the same thing: ⎛ RB ⎞ Vout (t ) = 20 sin ( 2π × 50t) × ⎜ ⎟ ⎝ RA + RB ⎠ It doesn’t matter that Vin now varies with time; at each point in time it’s still just multiplied by the factor RB RA + RB . • (d) Clearly the ratio Vout / Vin (the gain of the circuit) depends only on the resistors, and is RB RA + RB . If RA = RB the gain is just RB RB = = 0 .5 RB + RB 2 RB . That is, the output voltage is exactly half the input voltage. Expressing this gain in dB, we get: 20 log10 (0.5) = -6.02 dB. • Notice that the answer is negative because the output voltage is smaller than the input voltage. ELEC166 Worked Examples Week 5 2 (Q6) Plot the frequency spectrum of the following signal. In your plot remember that the ratio of amplitudes of different frequency components is reflected in the frequency spectrum. Write down an equation describing the signal as a function of time. 2 1.5 1 0.5 Voltage (V) 0 -0.5 -1 -1.5 -2 0 2 4 6 Time (ms) 8 ELEC166 Worked Examples Week 5 10 12 3 • Look at the signal carefully. If we draw in the dotted line as shown…. 2 1.5 1 0.5 Voltage (V) 0 -0.5 -1 -1.5 -2 0 2 4 6 Time (ms) 8 10 12 ..you can see that the original signal is made up of the sum of the following two sine waves, with peak amplitudes 1 and 0.5 volts: 2 1.5 1 0.5 Voltage (V) 0 -0.5 -1 -1.5 -2 0 2 4 6 Time (ms) 8 10 ELEC166 Worked Examples Week 5 12 4 • So these two sine waves are the frequency components of the signal. • From the graph, we can estimate their periods (T) – approximately 6.3 ms and 0.63 ms (notice that ten cycles of the smaller component fit in one cycle of the larger). • Since f = 1/ T the frequencies are thus: 1 1 = 1580 Hz = 158 Hz −3 −3 and 0.63 × 10 sec 6.3 × 10 sec • So the spectrum looks like this: Peak amplitude (volts) 1 0.5 0 158 Frequency (Hz) 1580 (Notice that the relative phase of the components is not shown in the spectrum) • The equation representing the signal will thus be v(t) = 1 sin (2π × 158 t) + 0.5 sin (2π × 1580 t) ELEC166 Worked Examples Week 5 5 For the spectrum below, sketch the individual frequency components as a function of time, and their sum. Use a time range of 0 – 20 ms for your plot. You may assume that all the components are of the form Asin(ωt) [rather than Asin(ωt + φ)] 3 Peak amplitude (volts) 2 1 0 50 100 200 Frequency (Hz) • Notice that the lowest frequency component is 50Hz, corresponding to a period of 1/50 sec, or 20ms. Thus the plot should cover one cycle of the 50 Hz waveform. • For the 100 Hz and 200 Hz components, there will be 2 and 4 cycles respectively. • The peak amplitudes at 50, 100 and 200 Hz are 3.0, 1.0 and 2.0 volts respectively. ELEC166 Worked Examples Week 5 6 • So the three components will look like this: 3 2 1 Voltage (V) 0 -1 -2 -3 0 0.005 0.01 0.015 0.02 0.015 0.02 Time (sec) …and their sum will look like this: 4 2 Voltage (V) 0 -2 -4 0 0.005 0.01 Time (sec) ELEC166 Worked Examples Week 5 7

Related documents