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Familiar with Canva but don’t know how to crack down on old designs? Canva’s user-friendly interface makes editing designs a piece of cake. Here is a step-by-step guide through the process!How to change templates in existing designs?Open an old design. If the design has multiple pages, then scroll through the pages and click anywhere inside the page you want to edit.From the side panel, select the Templates tab. A wide variety of templates with different designs appears. Select the design of your choice. You can also search for template using keywords in the search bar.Select the design you want to ... Read More
You’re launching a start-up or a new business, trying to make an impact. The first step to establishing your brand and its identity is choosing the most suitable color and fonts. It is the first impression that your customers get of your brand.Brand image is vital because you don’t want to convey a wrong message. It is key to understanding and educating your brand identity, a representation of how people wish to perceive you.What is Canva’s Brand Kit?In simple terms, brand kits are critical tools that organizations of all sizes and types need to build a cohesive brand. It is ... Read More
Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of Unit Impulse FunctionThe unit impulse function is defined as, $$\mathrm{\delta(t)=\begin{cases}1 & for\:t=0 \0 & for\:t ≠ 0 \end{cases}}$$If it is given that$$\mathrm{x(t)=\delta(t)}$$Then, from the definition of Fourier transform, we have, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}\delta(t)e^{-j\omega t}dt}$$As the impulse function exists only at t= 0. Thus, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}\delta(t) e^{-j\omega t}dt=\int_{−\infty}^{\infty}1\cdot e^{-j\omega t}dt=e^{-j\omega t}|_{t=0}=1}$$$$\mathrm{\therefore\:F[\delta(t)]=1\:\:or\:\:\delta(t) \overset{FT}{\leftrightarrow}1}$$That is, the Fourier transform of a unit impulse function is unity.The magnitude and phase representation of the Fourier transform of unit impulse function are as follows −$$\mathrm{Magnitude, |X(\omega)|=1;\:\:for\:all\:\omega}$$$$\mathrm{Phase, \angle X(\omega)=0;\:\:for\:all\:\omega}$$The graphical representation of the ... Read More
Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of Two-Sided Real Exponential FunctionLet a two-sided real exponential function as, $$\mathrm{x(t)=e^{-a|t|}}$$The two-sided or double-sided real exponential function is defined as, $$\mathrm{e^{-a|t|}=\begin{cases}e^{at} & for\:t ≤ 0\e^{-at} & for\:t ≥ 0 \end{cases} =e^{at}u(-t)+e^{-at}u(t) }$$Where, the functions $u(t)$ and $u(-t)$ are the unit step function and time reversed unit step function, respectively.Now, from the definition of Fourier transform, we have, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}e^{-a|t|}e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}[e^{at}u(-t)+e^{-at}u(t)]e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{0}e^{at}e^{-j\omega t}dt+\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{-\infty}^{0}e^{(a-j\omega)t}dt+\int_{0}^{\infty}e^{-(a+j\omega)t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{\infty}e^{-(a-j\omega)t}dt+\int_{0}^{\infty}e^{-(a+j\omega)t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\left[\frac{e^{-(a-j\omega)t}}{-(a-j\omega)}\right]_{0}^{\infty}+\left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty}=\left[\frac{e^{-\infty}-e^{0}}{-(a-j\omega)} \right]+\left[\frac{e^{-\infty}-e^{0}}{-(a+j\omega)} \right]}$$$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{a-j\omega}+\frac{1}{a+j\omega}=\frac{2a}{a^{2}+\omega^{2}}}$$Therefore, the Fourier transform of a two-sided real exponential function is, $$\mathrm{F[e^{-a|t|}]=X(\omega)=\frac{2a}{a^{2}+\omega^{2}}}$$Or, it can also be represented as, $$\mathrm{e^{-a|t|}\overset{FT}{\leftrightarrow}\frac{2a}{a^{2}+\omega^{2}}}$$Magnitude ... Read More
Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{x(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t }dt}$$Fourier Transform of Sine FunctionLet$$\mathrm{x(t)=sin\:\omega_{0} t}$$From Euler’s rule, we have, $$\mathrm{x(t)=sin\:\omega_{0} t=\left[\frac{ e^{j\omega_{0} t}- e^{-j\omega_{0} t}}{2j} \right]}$$Then, from the definition of Fourier transform, we have, $$\mathrm{F[sin\:\omega_{0} t]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}sin\:\omega_{0}\: t\: e^{-j\omega t}dt}$$$$\mathrm{ \Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}\left[ \frac{e^{j\omega_{0} t}-e^{-j\omega_{0} t}}{2j}\right] e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2j}\left[ \int_{−\infty}^{\infty}e^{j\omega_{0} t}e^{-j\omega t} dt-\int_{−\infty}^{\infty} e^{-j\omega_{0} t}e^{-j\omega t} dt\right]}$$$$\mathrm{=\frac{1}{2j}\{F[e^{j\omega_{0} t}] -F[e^{-j\omega_{0} t}]\}}$$Since, the Fourier transform of complex exponential function is given by, $$\mathrm{F[e^{j\omega_{0} t}]=2\pi\delta(\omega-\omega_{0})\:\:and\:\:F[e^{-j\omega_{0} t}]=2\pi\delta(\omega+\omega_{0})}$$$$\mathrm{ \therefore\:X(\omega)=\frac{1}{2j}[2\pi\delta(\omega-\omega_{0})-2\pi\delta(\omega+\omega_{0})]}$$$$\mathrm{\Rightarrow\:X(\omega)=-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$Therefore, the Fourier transform of the sine wave is, $$\mathrm{F[sin\:\omega_{0}\:t]=-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$Or, it can also be represented as, $$\mathrm{sin\:\omega_{0}\:t\overset{FT}{\leftrightarrow}-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$The graphical representation of the sine function with ... Read More
Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)= \int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of One-Sided Real Exponential FunctionA single-sided real exponential function is defined as, $$\mathrm{x(t)=e^{-a t}u(t)}$$Where, $u(t)$ is the unit step signal and is defined as, $$\mathrm{u(t)=\begin{cases}1 & for\:t≥ 0 \0 & for\:t < 0 \end{cases}}$$Then, from the definition of Fourier transform, we have, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}e^{-at}u(t)e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{\infty}e^{-(a+j\omega)t} dt=\left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)} \right]_{0}^{\infty}}$$$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{-(a+j\omega)}[e^{-\infty}-e^{0}]=\frac{0-1}{-(a+j\omega)}=\frac{1}{a+j\omega}}$$Therefore, the Fourier transform of a single-sided real exponential function is, $$\mathrm{F[e^{-at}u(t)]=\frac{1}{a+j\omega}}$$Or, it can also be represented as, $$\mathrm{e^{-at}u(t)\overset{FT}{\leftrightarrow}\frac{1}{a+j\omega}}$$Magnitude and phase representation of the Fourier transform of a single-sided real exponential function The Fourier transform of the ... Read More
Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as,$$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of Signum FunctionThe signum function is represented by $sgn(t)$ and is defined as$$\mathrm{sgn(t)=\begin{cases}1 & for\:t>0\-1 & for\:t
When starting out with the idea of a new brand, it's important to keep an eye on the legal side of things. When designing with Canva which provides you such a wide array of images, elements and features to work with, it can always be confusing to figure out what is free and legal to use. So, in here is a small article to help you deal with the question: "Are Canva images copyright free?"Canva – What is allowed and what is not?All images, videos and music that is available on Canva can be used both commercially and non-commercially. Canva ... Read More
Feeling lost while trying to find your old designs on Canva? For a new user, navigating your way through Canva can be a bit challenging, owing to the vast variety of features available on the platform. So, if you are someone who could use some help to access your old designs, then this is the place to visit. Here is a guide to help you out.Accessing previously created designs on CanvaFrom Canva homepage, select the All your designs option.This leads to a page with all your old designs. Scroll through the page each of the designs have a thumbnail you ... Read More
New to the world of blogging? Want to kick-start with some amazing graphics to enhance your visual aesthetic? Canva is the place for you. But a question that bugs most bloggers new to Canva is – "Can free templates be used on a monetized blog?" Let's discuss....Can we use free templates in a monetized blog?With reference to Canva’s license agreement, all images, videos, and music that is available on Canva can be used both commercially and noncommercially. So, yes, free templates can be used on a monetized blog as long as some kind of personal input in the form of ... Read More