ADC (Frequency Response)

Corner frequency

when imaginary part = jw only (otherwise a linear factor), if a transfer function is (jwRC + 1), the corner frequency is 1/RC

How do you obtain the low frequency asymptote?

the value of the lowest power of (jw) in the numerator divided by the lowest power in the denominator

How do you obtain the high frequency asymptote?

the value of the highest power of (jw) in the numerator divided by the highest power in the denominator

How do you find the slope in a magnitude response graph?

the power of (jw) will indicate the slope’s gradient, this is due to the magnitude response being a logarithmic graph

How do you determine the starting points and end points of a magnitude response graph?

Find the low frequency asymptote and high frequency asymptote expressions, turn their values into decibels (20log(v2/v1)). This assumes that the LFA and HFA are real values. For values with a power of (jw), graphing will be required (using logarithm and corner frequency).

What is special about the intersection of a Low Frequency Asymptote and High Frequency Asymptote?

The point of intersection is usually the corner frequency.

How do you plot magnitude response for a transfer function with multiple linear factors of (jw)?

Substitute linear factors with “w” based on a range involved w, and obtain modulus of resulting function.

For instance, with corner frequencies 1, 4, 12, 50 or (jw+1), (jw+4), (jw+12), (jw+50), between ranges 1 < w < 4, substitute w into (jw+1) in the main transfer function. For linear factors like (jw+12), (jw+50), a value of w in 1 < w < 4 is comparatively smaller to the real value in their linear factors, so these linear factors can be rounded to being just real (i.e. (jw+12) → (12)).

If there is a lone factor of (jw), that means for w < 1, (jw) is also replaced with w. For 1 < w < 4, (jw) is also substituted with w, this applies to any other linear factors.

The resulting answers of substitution will equal to an expression of a power of w. The power of w will determine the gradient of a slope between its relevant range (i.e. w < 1, 1 < w < 4).

How do you find low frequency and high frequency asymptotes for phase response?

Factorise transfer function expression so that it is made of linear factors of (jw).

For low frequency, (w « (linear factor)), so approximate imaginary (jw) as 0.

For high frequency, (w » (linear factor)), so approximate real linear factor as 0.

How do you find the starting and end points for phase response?

Find the low frequency and high frequency asymptote transfer function expressions and use arctan(imaginary/real). Whatever expression comes out is the starting (low frequency) and end (high frequency) point of the phase response.

Phase change is typically π/4 per decade (per 10* of rad/s)

How do you plot phase response for a transfer function of multiple linear factors?

List all corner frequencies (real part of each linear factor) and label them + or - according to their location on a fraction. + for numerator corner frequencies, - for denominator corner frequencies.

For every corner frequency, separate each of them into a decade range that encompasses the corner frequency. (e.g. for corner frequency 1^-, we have 0.1^- and 10^+)

Rearrange these decade values into ascending order (i.e. 0.1^-, 0.4^-). Calculate the distance (in rad/s) between them using log(w2/w1). This will give you the horizontal distance between each respective decade value.

± signs indicate the direction of “phase change”, + represents an increase in phase change (so +π/4), - represents a decrease in phase change (-π/4). If two decade values are opposite in sign to one another, they cancel the phase change of each other when one goes to the other. (i.e. from 0.4^-, gradient is -π/4 per decade, and when it reaches 1.2^+, gradient is now (-π/4 + π/4 = 0) per decade.

To find the change in vertical distance (or phase) between each respective decade value, multiply the horizontal distance between each respective decade value by the gradient of the slope of the decade interval involved.