Convolution Flashcards
if there is a linear system p(t) = q(t), what is the differentiation of p(t) equal to
d/dt[p(t)] = d/dt[q(t)]
if there is another linear system f(t) = y(t), what is the sum of f(t) and p(t) equal to
f(t) +p(t) = q(t) + y(t)
what does 5f(t) + 3p(t) equal
5f(t) + 3p(t) = 5y(t) + 3q(t)
what does a basic unit step function H(t) look like on a graph
- the line stays at H(t) = 0 for t < 0
- then when t = 0 it instantly shoots up to H(t) = 1
- and the line remains at H(t) = 1 forever
what does a delta / impulse function δ(t) look like on a graph
a vertical spike at a given x-axis value
what is the area of an impulse function spike
1
what is the derivative of a step function or step response
an impulse function or impulse response
what is the response of a function in this case
- the less ideal, more realistic version of the function’s behaviour
- because step and impulse functions act in very short time periods they are ‘modelled’ to have instantaneous reactions
- but step / impulse responses show the more ‘capacitor-like’ behaviour of these movements
if you have an impulse function δ(t-b), where is the spike
at t = b
if you have an arbitrary function f(t), what is the product of δ(t-b)*f(t)
δ(t-b)*f(t) = f(b)
what is the important trick to remember when evaluating whether the integration of the product of an arbitrary function and an impulse function is automatically 0
- work out the value of t in the impulse function bracket to get δ(0)
- if this value of t does not lie within the integration boundaries, the integration is automatically 0
if the value of t lies within the integration boundaries, what is another trick for making the integration easy
- simply input that value of t into the arbitrary function and thats your answer
- no actual integration is needed
for finding the step response to a differential equation that equals f(t), what is the first step
set f(t) = 1 and find the general solution
what is the second step
- set the boundary conditions y(0) = 0 and solve for the constant
- if its a second order system set y’(0) = 0 too
what is the third step
- write in the curly brackets that y = 0 for t < 0
- and y = the solution found for t => 0
if you have a system input composed of impulses f(t) = 3σ(t - 1) + 4σ(t - 2), how many impulse responses are there and at what times
- 3 impulse responses at t = 1
- 4 impulse responses at t = 2
what is the convolution integral formula
y(t) = int[g(t - T)*f(T) dt] from limits t to -∞
what is the difference between t and T (actually tau but the symbol doesn’t exist apparently)
- t is the time as it relates to the output of the system y(t)
- T is the time as it relates to the input of the system f(T)
- also t is not the integration variable, T is
where do you get the value for f(T)
- it is the value(s) in the curly brackets for the input
- if those values are a function themselves like sin(t), replace the t with a T when adding it to the integral
what do you do if f(t) is a discontinuous function
- work out the convolution integral for each section of the function
- piece them together in the end
what is the overall step by step process for convolution when starting with a differential equation
- get the step response by equating it to 1 and using y(0) = y’(0) = 0
- differentiate the step response to get the impulse response g(t)
- IF the step response is the input, use the convolution integral using the step and impulse response
- if the input is simply given, use that in the convolution integral instead
- the answer for both is the output y(t)
if you get a question like “find the impulse response of … = f(t) and hence find the output when input f(t) = H(t)*e^t, what is the important thing to remember about what this means
- it means the required input f(T) in the convolution integral is the e^t part (actually e^T in the integration)
- so if it just says f(t) = H(t), your input is just 1
what is the name of the other method we can use for evaluating convolution integrals when g(t) is too complicated
the alternative convolution integral
how do you set up the alternative convolution integral (this makes no sense as to why it works but anyway)
- set T(2) = t - T(1)
- T1 is the tau we’ve been using this whole time and T2 is a “‘new’ tau but has no subscripts, they look exactly the same
- sub into the convolution integral so f(T) = t - T and g(t - T) = g(T)
- giving y(t) = int[f(t - T)*g(T) dT] from limits t to 0
