A band limited noise of (0.5 volt^2) power is added to sinusoidal signal with SNR (20 dB ) .The power of sine wave signal is www www O 100 volt^2 O 10 volt^2 O 5 volt^2 50 volt^2 O O
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- 1. Use the Gram-Schmit procedure to obtain an orthogonal basis for the set of waveforms shown in Figure 1. 0.4T T. 0.4T 0.8T 0.6TQ.2) Determine all values of resistors and capacitor for the triangular waveform generator whose output alternates between +10Vpeak and -10V peak when the input is a 10-Hz square wave that alternates between +2V and -2V. The input resistance to the generator must be at least 15 kQ. you may assume that there is no de component or offset in the input.A clipper is a device that removes either the positive half (top half) or negative half (bottom half), or both positive and negative halves of the input AC signal. In other words, a clipper is a device that limits the positive amplitude or negative amplitude or both positive and negative amplitudes of the input AC signal. In some cases, a clipper removes a small portion of the positive half cycle or negative half cycle or both positive and negative half cycles. In the below circuit diagram, the positive half cycles are removed by using the series positive clipper. QUESTION: In your own opinion. Why do we need to clip a certain amount of voltage in positive or negative or on both sides? What is the benefit of that in our devices or circuits in doing such thing?
- Determine the rms current in the figure for Vrms = 16 V and C = 74µF V f=10 kHzFrom step1, how did you get that 50+(80*j60/80+j60) = 78.8+j38.4? also in step 2, how did you get that 5/(0.0125-0.0167) = 143.63 + j191.89?In DC choppers the wave forms for input and output voltages are respectively a. both discontinuous b. discontinuous,continuous c. both continuous d. continuous,discontinuous
- A pure sinusoidal current is being rectified. For the given maximum value of half wave rectified current is 50 A, then the rms value of full wave rectification will be 50 (a) 100 (b) - - (c) 100 A (d) 70.7 AA pure sinusoidal current is being rectified. For the given maximum value of half wave rectified current is 50 A, then the rms value of full wave rectification will be 50 (a) A (b) 100 - A TC (c) 100 A (d)70.7 AFor the below sinusoidal voltage trace displayed by the CRO, the time/cm switch is on 5ms/cm and the volt/cm switch is on 3 V/cm. a) the frequency will bel Hz b) the amplitude will be V.
- The peak distance of a sinusoidal waveform displayed on a C.R.O. screen is 6 cm and the 'volts/cm' switch is on 33 V/cm. The peak to peak voltage is given by O 396.00 V O 5.50 V O 139.99 V O 99.00 VIn an experiment, the function generator is adjusted to generate a square voltage at a certain frequency. This voltage is displayed on an oscilloscope and the reading is 16 V peak-to-peak. The rms value of this voltage is .What will be the rms value of this voltage if the frequency is doubled? Select one: O 16 Vrms 32 Vrms O 8 Vrms . 8 Vrms ..... O 5.657 Vrms 5.657 Vrms ....... O 8 Vrms . 16 Vrms O 16 Vrms 8 Vrms ......A certain impedance is given by 1012° N. Does the current lead or lag the voltage, and by how many degrees? Fill in the blanks. The current the voltage by