The derivative of a triangle-wave is a square-wave. The illustration below shows a triangle- wave having...

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The derivative of a triangle-wave is a square-wave. The illustration below shows a triangle- wave having an amplitude A and a period T. On a line segment with a positive slope, the triangle-wave value changes by 2A (peak to peak) over a time span of T/2. The slope is 4A/T. This is also the derivative for this line segment (with the positive slope). For a line segment with a negative slope, the derivative is -4A/T. The derivative of this triangle-wave is therefore the square-wave shown below. (We won't worry about the derivative at the peaks of the triangle- wave.) 2A 44/T -44/T T/2 In a capacitor v and i are related by i = C dv dt where C is the capacitance in farads, v is in volts and i is in amperes. In the special case where v is a constant (DC), then i is 0: no current flows when the voltage across a capacitor is DC. In other words, a capacitor looks like an open circuit to DC. We wish to know the current i through a capacitor (having capacitance C) if the voltage v across it is a triangle-wave having amplitude A and period T. We can take advantage of the math given above. The derivative dv/dt is a square-wave with amplitude 4A/T. Therefore, i is a square- wave with amplitude 4AC/T. Exercise: A triangle-wave voltage with amplitude 2 V and frequency 1 kHz is applied across a capacitor. The capacitance is 0.1 µF. Based on the mathematics discussed above, we can say that the current will be a square-wave. What is the amplitude of the square-wave current? The derivative of a triangle-wave is a square-wave. The illustration below shows a triangle- wave having an amplitude A and a period T. On a line segment with a positive slope, the triangle-wave value changes by 2A (peak to peak) over a time span of T/2. The slope is 4A/T. This is also the derivative for this line segment (with the positive slope). For a line segment with a negative slope, the derivative is -4A/T. The derivative of this triangle-wave is therefore the square-wave shown below. (We won't worry about the derivative at the peaks of the triangle- wave.) 2A 44/T -44/T T/2 In a capacitor v and i are related by i = C dv dt where C is the capacitance in farads, v is in volts and i is in amperes. In the special case where v is a constant (DC), then i is 0: no current flows when the voltage across a capacitor is DC. In other words, a capacitor looks like an open circuit to DC. We wish to know the current i through a capacitor (having capacitance C) if the voltage v across it is a triangle-wave having amplitude A and period T. We can take advantage of the math given above. The derivative dv/dt is a square-wave with amplitude 4A/T. Therefore, i is a square- wave with amplitude 4AC/T. Exercise: A triangle-wave voltage with amplitude 2 V and frequency 1 kHz is applied across a capacitor. The capacitance is 0.1 µF. Based on the mathematics discussed above, we can say that the current will be a square-wave. What is the amplitude of the square-wave current?

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