Convolution integral, Fourier and Laplace transform

Convolution integral, Fourier and Laplace transform Introduction In this paper, I tried to explain why we need Fourier and Laplace transform when analyzing continuous time systems. I started with the definition of continuous signals and LTI systems, then by exploiting linearity and time invariance properties of LTI system, I elucidated convolution by showing where actually it comes from. Next, Fourier and Laplace transforms were introduced. I also shed a light on key propertiy of both Laplace and Fourier transform which states that transform of convolution of two signals in time domain, turns to be multiplication in both frequency and s domain. Moreover, I introduced two important applications of this property in filtering and stability analysis. Continuous Signals and LTI Systems Signals whose value is varying continuously with time are considered as continuous signals. Examples of continuous signals may encompass temperature of ambient air, level of liquid in a tank, speech signal etc. Systems which produce continuous output when excited by continuous input are referred as continuous systems. Spring-mass system, simple filters constructed by resistor, capacitor and inductor are typical examples of continuous systems. In this paper, we will focus on Linear Time Invariant Systems. Linearity implies that superposition is met which can be explained like: if the output of system for an input 𝑥1 (𝑡) is 𝑦1 (𝑡) and for 𝑥2 (𝑡) is 𝑦2 (𝑡), then output for the input 𝑎𝑥1 (𝑡) + 𝑏𝑥2 (𝑡) will be 𝑎𝑦1 (𝑡) + 𝑏𝑦2 (𝑡). a and b are constant real numbers. Another important aspect here is time-invariance, which states that if input 𝑥(𝑡) is delayed by some amount, then output 𝑦(𝑡) will also be delayed by the same amount. It can be shown mathematically as following: 𝑥(𝑡) → 𝑦(𝑡) 𝑥(𝑡 − 𝑑) → 𝑦(𝑡 − 𝑑) Where d denotes an amount of delay in seconds. It is important to understand a relationship between an input to the system and produced output. Actually there is very simple relationship between input and corresponding output which we call convolution integral. In the next phase, we will consider convolution integral in continuous systems. Convolution integral in continuous time systems First, we have to represent our input 𝑥(𝑡) as linear combinations of shifted Dirac Delta functions in order to understand where convolution come from. Now let us define another function 𝛿∆ (t), whose graph is like following: Value of 𝛿∆ (t) is equal to 1 ∆ when 0 <= 𝑡 <= ∆, and is zero otherwise. By definition, area under 𝛿∆ (t) is always equal to unity independent of ∆. Similarly, we can show 𝛿∆ (t-∆) as in following graph: We can approximate our signal x(t) as linear combinations of 𝛿∆ as following: 𝑥(𝑡) can be approximated as: 𝑥(𝑡) ≅ ⋯ + 𝑥(−∆)𝛿∆ (𝑡 + ∆)∆ + 𝑥(0)𝛿∆ (𝑡)∆ + 𝑥(∆)𝛿∆ (𝑡 − ∆)∆ + ⋯ Above equation means that, the value of approximated function between 𝑘∆<= 𝑡 <= (𝑘 + 1)∆ will be equal to the value of 𝑥(𝑘∆). In discrete sum form: +∞ 𝑥(𝑡) ≅ ∑ 𝑥(𝑘∆) 𝛿∆ (𝑡 − 𝑘∆)∆ 𝑘=−∞ Our approximation will get more precise as ∆ becomes smaller, and limit of this sum, as ∆ approaches to zero will be exact reconstruction of 𝑥(𝑡), which is integral by definition. So: +∞ +∞ 𝑥(𝑡) = lim ∑ 𝑥(𝑘∆) 𝛿∆ (𝑡 − 𝑘∆)∆ = ∫ 𝑥(𝜏) 𝛿(𝑡 − 𝜏)𝑑𝜏 ∆→∞ 𝑘=−∞ −∞ Above equation implies that, any continuous time signal 𝑥(𝑡) can be shown as the continuous sum of shifted and scaled impulses. Therefore, if we know the response of system for single impulse, then we can calculate the response of the system for any input since our system is linear and time-invariant. An output of LTI system for 𝛿(𝑡) is called an impulse response of system and is denoted by ℎ(𝑡). If we know ℎ(𝑡), then we know everything about our system. According to linearity and time invariance of system, the output of our system for any input 𝑥(𝑡) can be calculated as a convolution integral as following: +∞ +∞ 𝑦(𝑡) = x(t) ∗ h(t) = ∫ ℎ(𝜏) 𝑥(𝑡 − 𝜏)𝑑𝜏 = ∫ 𝑥(𝜏) ℎ(𝑡 − 𝜏)𝑑𝜏 Fourier transform −∞ −∞ Now let us consider 𝑥(𝑡) equal to 𝑥(𝑡) = 𝑒 𝑖𝜔𝑡 , which is complex number. 𝑒 𝑖𝜔𝑡 is equal to cos( 𝜔𝑡) + 𝑖𝑠𝑖𝑛(𝜔𝑡). Since ℎ(𝑡) is real, the output will also be complex number with real and imaginary part. The reason why I chose 𝑒 𝑖𝜔𝑡 as an input to the system is that, by obtaining response of the system for 𝑒 𝑖𝜔𝑡 , we actually obtain the response of the system for cos( 𝜔𝑡) and 𝑠𝑖𝑛(𝜔𝑡) which is real and imaginary part of complex output, respectively. +∞ +∞ +∞ 𝑦(𝑡) = ∫ ℎ(𝜏) 𝑥(𝑡 − 𝜏)𝑑𝜏 = ∫ ℎ(𝜏) 𝑒 𝑖𝜔(𝑡−𝜏) 𝑑𝜏 = ∫ ℎ(𝜏) 𝑒 𝑖𝜔𝑡 𝑒 −𝑖𝜔𝜏 𝑑𝜏 −∞ −∞ Since 𝑒 𝑖𝜔𝑡 does not depend on 𝜏 we can rewrite 𝑦(𝑡) as: +∞ +∞ −∞ +∞ 𝑦(𝑡) = 𝑒 𝑖𝜔𝑡 ∫ ℎ(𝜏) 𝑒 −𝑖𝜔𝜏 𝑑𝜏 = 𝑥(𝑡) ∫ ℎ(𝜏) 𝑒 −𝑖𝜔𝜏 𝑑𝜏 −∞ −∞ It is clear that ∫−∞ ℎ(𝑡) 𝑒 −𝑖𝜔𝑡 𝑑𝑡 is a complex number, whose magnitude and phase depends on the value of 𝜔 for given system. We can denote this complex number like 𝐻(𝑖𝜔) which can be shown as |𝐻(𝑖𝜔)|〈𝜑(𝜔)〉 in phasor form where |𝐻(𝑖𝜔)| and 〈𝜑(𝜔)〉 are magnitude and phase associated with angular frequency 𝜔. Similiarly we can denote 𝑒 𝑖𝜔𝑡 in phasor notation as 1 〈𝜔〉. When we multiply two complex numbers, magnitudes will be multiplied and phases will be added like following: +∞ 𝑦(𝑡) = 𝑒 𝑖𝜔𝑡 ∫ ℎ(𝜏) 𝑒 −𝑖𝜔𝜏 𝑑𝜏 = 1 〈𝜔〉 |𝐻(𝑖𝜔)| 〈𝜑(𝜔)〉 −∞ 𝑦(𝑡) = |𝐻(𝑖𝜔)|〈𝜔 + 𝜑(𝜔)〉 = |𝐻(𝑖𝜔)|cos((𝜔 + 𝜑(𝜔))𝑡) + 𝑖|𝐻(𝑖𝜔)|sin((𝜔 + 𝜑(𝜔))𝑡) Therefore, output for cos( 𝜔𝑡) and sin( 𝜔𝑡) will be real and imaginary part of complex output We actually proved that system does not affect the frequency of input signal, what changes are only magnitude and phase of sinusoid. +∞ ∫−∞ ℎ(𝑡) 𝑒 −𝑖𝜔𝑡 𝑑𝑡 Is actually Fourier transform of ℎ(𝑡). Since we have impulse response ℎ(𝑡), we can calculate its Fourier transform. In general Fourier transform of x(t) can be shown as: +∞ 𝑋(𝑖𝜔) = ∫ 𝑥(𝑡) 𝑒 −𝑖𝜔𝑡 𝑑𝑡 −∞ We can actually separate this formula to its real and imaginary part like: +∞ +∞ 𝑋(𝑖𝜔) = ∫ 𝑥(𝑡) cos( 𝜔𝑡) 𝑑𝑡 + 𝑖 ∫ 𝑥(𝑡) sin( 𝜔𝑡) 𝑑𝑡 −∞ −∞ We have to emphasize that, for some signals, (for example, 𝑥(𝑡) = 𝑒 2𝑡 𝑢(𝑡) or 𝑥(𝑡) = 𝑡 2 𝑢(𝑡) ) Fourier Transform will no converge, which is main drawback of FT. There are some values of 𝜔 that makes Fourier transform equal to zero, in this case input containing these frequencies will generate zero output. An important point here is that Fourier transform of convolution 𝑥(𝑡) ∗ ℎ(𝑡) will be multiplication 𝑋(𝑖𝜔)𝐻(𝑖𝜔). This is very important property that makes Fourier transform useful in filter design. +∞ 𝑦(𝑡) = ∫ ℎ(𝜏) 𝑥(𝑡 − 𝜏)𝑑𝜏 −∞ Let us get Fourier transform from both sides, Fourier transform of 𝑦(𝑡) is indicated as 𝑌(𝑖𝜔): 𝑌(𝑖𝜔) = ∫ +∞ −∞ {∫ +∞ −∞ Let’s denote above equation like following: Arranging terms: By manupilation: 𝑌(𝑖𝜔) = ∫ +∞ −∞ 𝑌(𝑖𝜔) = ∫ {∫ +∞ −∞ 𝑌(𝑖𝜔) = {∫ −∞ +∞ −∞ {∫ +∞ ℎ(𝜏)𝑥(𝑡 − 𝜏)𝑑𝜏} 𝑒 −𝑖𝜔𝑡 𝑑𝑡 ℎ(𝜏)𝑥(𝑡 − 𝜏)𝑑𝜏} 𝑒 −𝑖𝜔𝑡 (𝑒 −𝑖𝜔𝜏 𝑒 𝑖𝜔𝜏 )𝑑𝑡 +∞ −∞ ℎ(𝜏)𝑥(𝑡 − 𝜏)𝑑𝜏} 𝑒 −𝑖𝜔(𝑡−𝜏) 𝑒 −𝑖𝜔𝜏 𝑑𝑡 𝑥(𝑡 − 𝜏)𝑒 −𝑖𝜔(𝑡−𝜏) 𝑑𝑡} {∫ We end up with the following result: +∞ −∞ ℎ(𝜏)𝑒 −𝑖𝜔𝜏 𝑑𝜏} = 𝑋(𝑖𝜔)𝐻(𝑖𝜔) 𝑌(𝑖𝜔) = 𝑋(𝑖𝜔)𝐻(𝑖𝜔) This equation is very important in order to understand how filter works. If an impulse response is a sinc function, then its Fourier transfrom will be rectangle. Now let us suppose that Fourier transform of 𝑥(𝑡) and ℎ(𝑡) is as following, then frequency spectrum of 𝑦(𝑡) can be obtained easily just by multiplying 𝑋(𝑖𝜔) and 𝐻(𝑖𝜔). 𝜔𝑐 here is cutoff frequency which is maximum frequency that filter let pass, frequencies more than 𝜔𝑐 are removed. 𝜔𝑁 is the highest frequency component existing in the 𝑥(𝑡), our filter will attenuate all frequencies which are greater than 𝜔𝑐 . The value of cutoff frequency 𝜔𝑐 and gain of filter surely depend on an impulse response of the system. Laplace transform Laplace transform is actually generalization of Fourier transform. In order to understand why LT is important in signals and systems, let’s consider 𝑥(𝑡) = 𝑒 𝑠𝑡 , where s is complex number 𝑠 = 𝜎 + 𝑖𝜔 Output of system will be again: +∞ +∞ 𝑦(𝑡) = ∫ ℎ(𝜏) 𝑥(𝑡 − 𝜏)𝑑𝜏 = 𝑦(𝑡) = ∫ ℎ(𝜏) 𝑒 −∞ −∞ 𝑠(𝑡−𝜏) 𝑑𝜏 = ∫ ℎ(𝜏) 𝑒 𝑠𝑡 𝑒 −𝑠𝜏 𝑑𝜏 Since 𝑒 𝑠𝑡 does not depend on 𝜏, we can rewrite y(t) as following: −∞ +∞ +∞ +∞ +∞ y(t) = 𝑒 𝑠𝑡 ∫ ℎ(𝜏) 𝑒 −𝑠𝜏 𝑑𝜏 = 𝑥(𝑡) ∫ ℎ(𝜏) 𝑒 −𝑠𝜏 𝑑𝜏 −∞ ∫−∞ ℎ(𝜏) 𝑒 −𝑠𝜏 𝑑𝜏 is a Laplace Transform of ℎ(𝑡). −∞ Laplace transform is actually not restricted to only impulse response, but very broad topic. When discussing Fourier Transform, we mentioned that FT will not converge for typical signals like 𝑥(𝑡) = 𝑒 2𝑡 𝑢(𝑡) or 𝑥(𝑡) = 𝑡 2 𝑢(𝑡), whose value is getting bigger as time goes. Therefore, it necessitates the introduction of generalized Fourier Transform, which is well-known Laplace transform. When calculating LT for the signal x(𝑡) , we multiply 𝑥(𝑡) by decaying exponential 𝑒 −𝜎𝑡 , and take its Fourier Transform. LT of signal x(𝑡) is calculated as an following integral where 𝑠 = 𝜎 + 𝑖𝜔 +∞ X(s) = ∫ 𝑥(𝑡) 𝑒 −𝑠𝑡 𝑑𝑡 −∞ By substituting 𝑠 = 𝜎 + 𝑖𝜔 , we get: +∞ +∞ X(s) = ∫ 𝑥(𝑡) 𝑒 −(𝜎+𝑖𝜔)𝑡 𝑑𝑡 = ∫ 𝑥(𝑡) 𝑒 −𝜎𝑡 𝑒 −𝑖𝜔𝑡 𝑑𝑡 −∞ −∞ When 𝜎 = 0, LT will reduce to FT, otherwise you can think of it as FT of 𝑥(𝑡) 𝑒 −𝜎𝑡 . The range of 𝜎 that makes LT to converge is called region of convergence (ROC). Let us consider one example regarding ROC: Suppose that 𝑥(𝑡) = 𝑒 −5𝑡 𝑢(𝑡), substituting 𝑥(𝑡) into LT formula we get: +∞ +∞ 0 0 𝑋(𝑠) = ∫ 𝑒 −5𝑡 𝑒 −𝜎𝑡 𝑒 −𝑖𝜔𝑡 𝑑𝑡 = ∫ 𝑒 −(5+𝜎)𝑡 𝑒 −𝑖𝜔𝑡 𝑑𝑡 Necessary condition for convergence here is 𝜎 > −5 , because otherwise LT will not converge. ROC for this example on s plane is shown in following diagram: ROC only depends on 𝜎, not 𝜔. The value of real signals will be equal to zero for negative time, therefore Laplace transform is reduced to following form: If 𝑥(𝑡) is zero for 𝑡 < 0, then LT is reduced to the following form: +∞ X(s) = ∫ 𝑥(𝑡) 𝑒 −𝑠𝑡 𝑑𝑡 0 Similar to Fourier transform, Laplace transform of convolution 𝑥(𝑡) ∗ h(t) will be multiplication namely X(s)H(s). You can derive this formula in the same way we did for FT. This is a key property of Laplace transform. +∞ 𝑦(𝑡) = ∫ ℎ(𝜏) 𝑥(𝑡 − 𝜏)𝑑𝜏 −∞ 𝑌(𝑠) = 𝑋(𝑠)𝐻(𝑠) As mentioned before 𝐻(𝑠) is LT of ℎ(𝑡). Therefore, if we are given 𝐻(𝑠), then it is possible to find ℎ(𝑡). Let us assume that: 𝐻(𝑠) = 2 5 + (𝑠 + 1) (𝑠 + 6) Values of s which makes denominator zero are called the poles of system (in this case 𝑠 = −1 and 𝑠 = −6 are poles of the system), while values of s making numerator zero are zeros of system (In this case 𝑠 = − 32 7 ) For system to be stable, all poles of transfer function must have negative real part. In order to justify this statement, we have to get ℎ(𝑡) from 𝐻(𝑠) which can be done by Inverse Laplace transform: ℎ(𝑡) = 5 𝑒 −𝑡 𝑢(𝑡) + 2 𝑒 −6𝑡 𝑢(𝑡) ℎ(𝑡) is response of system for 𝛿(𝑡), it is clear that response goes to zero as time goes to infinity. Finally, we can deduce that an output for bounded input is also bounded. +∞ 𝑥(𝑡) = 𝛿(𝑡) − −−→ 𝑦(𝑡) = ℎ(𝑡) +∞ 𝑥(𝑡) = ∫ 𝑥(𝜏) 𝛿(𝑡 − 𝜏)𝑑𝜏 − −−→ 𝑦(𝑡) = ∫ 𝑥(𝜏) ℎ(𝑡 − 𝜏)𝑑𝜏 −∞ −∞ Author: Kamil Budagov Position: Process Automation Engineering Student at Baku Higher Oil School