a phase shifted sine wave of a different amplitude.
A sine wave is the graph of y = sin(x). It demonstrates to cyclic nature of the sine function.
Because if one applies a step wave or a sine wave, we will not be able to observe the transcient response. In case of square wave it would have already happened at the start and hence we wont be able to observe the transcient part in the response, taking output,say in a CRO.In case of a sine wave , since the voltages are constantly changing it would not be possible to observe a proper transcient response and one will need to wait atleast until the settling time,to change the voltage of input.
The differential of the sine function is the cosine function while the differential of the cosine function is the negative of the sine function.
The period of an 8000 Hz sine wave is 0.125 milliseconds. (1/8000)
A sine wave is a periodic function and, by suitably adjusting the argument of the sine function, can be made to fit a wide functions with different frequencies.
If you feed a sine wave through an amplifier that isn't exactly linear, the output will be distorted, not a pure sine wave. Distortion is the defect where the output from a device does not mirror the input.
Oh, dude, if a sine wave is the input to a NOT gate, the output will be the inverted sine wave. It's like flipping the wave upside down, you know? So, if the input is high, the output will be low, and vice versa. It's just how the NOT gate rolls, man.
When a low pass filter is used with a sine wave input, the output is also a sine wave. The output will be reduced in amplitude and phase shifted when the frequency is high, but it is still a sine wave. This is not the case for square or triangular wave inputs. For non-sinusoidal inputs the circuit is called an integrator.
It doesn't. It can produce any waveform if you feed the integral of the desired waveform into the differentiator's input.
If a sine wave is applied to a rectifier, and the sine wave is strictly AC (no DC offset), the output will be 1/2 the wave - it will be clipped near zero, as the diode prevents reverse voltages. So the output will NOT be a perfect sine wave.
The output frequency of the half-wave rectifier will be 60 Hz if the input is a 60 Hz sine wave.One cycle of the input will include the positive going and the negative going portions of the sine wave. The output will have either the positive going or negative going half of the input wave, and will have no output during the other half of the input sine wave when the diode is reverse biased. What that output will look like on an oscilloscope is half a wave and then a "flat spot" where there is no output (owing to the diode being reversed biased). Let's keep going.The frequency of a signal is the number of cycles of the signal per second. Further, we know that in a waveform, one cycle occurs when the wave goes through all of the changes it must go through to, shall we say, get back to where it started. In the half-wave output, the signal goes through half of the input wave, and then the voltage sits at zero. That means that one output cycle consists of that voltage excursion, and that period during which the diode is back biased. So the time for one complete cycle of the output is the same as the time for one complete cycle of the input. Thus, a 60 Hz input signal (that sine wave) will give us a half-wave rectified 60 Hz output signal.
Try the mathematics and you will see how.For f(x) = ∫x dt, where x is a square wave function, f(x) will be a triangle wave function.Also try what happens where x is a triangle wave function!
The frequency of a full-wave rectifier is double that of the input, if the input is a sine wave or triangle wave. If the input is a square wave, the output is DC. If the input is a sawtooth wave, the output is a triangle wave of the same frequency.
infinite positive impulse wave on each leading edge, infinite negative impulse wave on each trailing edge. the rest of the time it is 0V. this assumes an ideal square wave, which I must assume as you did not give rise/fall times or edge slew rates.
You need a differentiator circuit, the simplest of which passes the input through a capacitor to the inverting terminal of a fedback op-amp. The R and C you choose to use depends on the frequency and gain of the signal you are trying to output. See the wikipedia article on operational amplifiers, and find the differentiator, not the differential amplifer (totally different)
square wave output without 180 deg shift...
If a square wave is used instead of a sine wave in a transformer, the output power will operate at a different frequency. This will produces varying levels of voltage and amperage based on the wave.